[tex] - 15 \ \textless \ - 4x - 3[/tex]That's the Math problem

Answers

Answer 1

EXPLANATION

Given the inequality

-15 < -4x - 3

Adding +4x to both sides:

-15 + 4x < -4x + 4x -3

Adding +15 to both sides:

-15 + 15 + 4x < -3 + 15

Simplifying:

4x < -15 - 3

Adding like terms:

4x < -15 - 3

Dividing both sides by 4:

x < -18/4

Simplifying the fraction:

x < -4.5

The solution is x<-4.5


Related Questions

The graph of an inequality has a closed circle at 4.3, and the ray moves to the right. What inequality is graphed?x > 4.3x ≥ 4.3x ≤ 4.3x < 4.3

Answers

From the question, we were told that the graph of an inequality has a closed circle at 4.3 and the ray moves to the right also.

We are to determined the inequality that is graphed from the options.

From what is seen, we want x to be greater than or equal to 4.3.

The closed circle tells us that it can be equal to 4.3. The ray to the right tells us that we are looking for numbers larger than 4.3.

So the inequality graphed is that of x is greater than or equal to 4.3

So the correct option is the second option which is x ≥ 4.3.

list all numbers from the given set that are

Answers

Part a

Natural numbers are:

[tex]\sqrt[]{25}[/tex]

because

[tex]\sqrt[]{25}=5[/tex]

Part b

whole numbers

[tex]0,\text{ }\sqrt[]{25}[/tex]

Part c

Integers

[tex]-9,0,\sqrt[]{25}[/tex]

Part d

rational numbers

[tex]\frac{3}{4},-9,0.6,0,8.5,\sqrt[]{25}[/tex]

Part e

Irrational numbers

[tex]\pi,\text{ }-\sqrt[]{2}[/tex]

Part f

real numbers

[tex]\frac{3}{4},-9,0.6,0,\pi,8.5,\sqrt[]{25},\text{ -}\sqrt[]{2}[/tex]

Geometry Question: Given segment EA is parallel segment DB, segment EA is congruent to segment DB, and B is the mid of segment AC; Prove: segment EB is parallel to segment DC (reference diagram in picture)

Answers

Construction: Join ED.

The corresponding diagram is given below,

According to the given problem,

[tex]\begin{gathered} AE=BD \\ AE\parallel BD \end{gathered}[/tex]

Since a pair of opposite sides are parallel and equal, it can be claimed that quadrilateral ABDE is a parallelogram.

Then, as a property of any parallelogram, it can be argued that,

[tex]\begin{gathered} AB=DE \\ AB\parallel DE \end{gathered}[/tex]

Given that B is the mid-point of AC,

[tex]\begin{gathered} AB=BC \\ AB\parallel BC \end{gathered}[/tex]

Combining the above two results,

[tex]\begin{gathered} BC=DE \\ BC\parallel DE \end{gathered}[/tex]

It follows that ABCD also forms a parallelogram.

Again using the property that opposite sides of a parallelogram are equal and parallel. It can be claimed that,

[tex]\begin{gathered} EB=DC \\ EB\parallel DC \end{gathered}[/tex]

Hence proved that segment EB is parallel to segment DC,

[tex]\vec{EB}=\vec{DC}[/tex]

In the first week of​ July, a record 1,040 people went to the local swimming pool. In the second​ week,125 fewer people went to the pool than in the first week. In the third​ week,135 more people went to the pool than in the second week. In the fourth​ week,322 fewer people went to the pool than in the third week. What is the percent decrease in the number of people who went to the pool over these four​ weeks?

Answers

By the concept of percentage there is 20% decrease in the number of people who went to the pool over these four​ weeks.

What is percentage?

A percentage is a statistic or ratio that is expressed as a fraction of 100 in mathematics. But even though the abbreviation "pct.", "pct.", and occasionally "pc" are also used, the percent symbol, "%," is frequently used to signify it. A % is a dimensionless number; there is no specific unit of measurement for it. %, a relative figure signifying hundredths of any amount. Since one percent (symbolized as 1%) is equal to one hundredth of something, 100 percent stands for everything, and 200 percent refers to twice the amount specified. A percentage is a figure or ratio that in mathematics represents a portion of one hundred. It is frequently represented by the sign "%" or just "percent" or "pct." For instance, the fraction or decimal 0.35 is comparable to 35%.

In July:

First week:

Number of people went to the local swimming pool

=1040

Second week:

110 fewer people went to the pool than in the first week

Number of people went to the local swimming pool

=1040 - 110

=930

Third week:

130 more people went to the pool than in the second week

Number of people went to the local swimming pool

=930 + 130

=1060

Fourth week:

228 fewer people went to the pool than in the third week

Number of people went to the local swimming pool

=1060 - 228

=832

Decrease in number of people over four week

= number of people in first week - number of people in fourth week

Decrease in number of people over four week

=1040 - 832

=208

Now, the percentage

=  20%

To know more about percentage ,visit:

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1.For a standard normal distribution, find:P(1.26 < z < 1.48)2.For a standard normal distribution, given:P(z < c) = 0.1288

Answers

Standard Normal Distribution

To find the cumulative probability of a Normal Distribution, we need to use some automated digital tool that makes the calculations for us, since it's a pretty complex formula.

We'll use an online tool and provide the results here.

a) P(1.26 < z < 1.48)

The procedure is: Find P(z < 1.48) directly from the tool. Find P(z < 1.48) also. Subtract both values.

P(z < 1.48) = 0.931

P(z < 1.26) = 0.896

Subtract the values above: 0.931 - 0.896 = 0.035. Thus:

P(1.26 < z < 1.48) = 0.035

b) Find c such that: P(z < c) = 0.1288

We need to use the inverse Normal Distribution, enter the probability and find the z-score: c = -1.132

Can you please help me out with a question

Answers

the figure is composed by a 4 triangles and a cube

to find the area of a triangle we need the base and height. the base is 15ft

to find the height we mut use the pithagorean theorem

h= height of the traingle

[tex]h^2=(15ft)^2+(7.5ft)^2[/tex]

resolving we have

[tex]h=\sqrt[\square]{281.25}\text{ = 16.78 aprox}[/tex]

and now he have all the measures

each triangle at the top has an area equal to

[tex]A=\frac{16.78ft\cdot15ft}{2}=127.78sq\text{ ft}[/tex]

now we multiply that by 4: 127.78sq ft*4=503.1 sq ft

for the bottom part, there are 5 squares of side 15ft

each square has an area = 15ft*15ft = 225 sq ft

multipliying that by 5: 225sqft*5=1125 sq ft

the total area is 1125 sq ft+503.1sqft=1628.1 sq ft rounded is 1628 sq ft

For the volume of the piramid, we use

[tex]V=\frac{1}{3}A\cdot h[/tex]

where A is the area of the base and h is the height

so volume of piramid:

[tex]V=\frac{1}{3}\cdot225\text{sqft}\cdot15ft=1125ft^3[/tex]

for the volume of the cube we multiply the side length 3 times:

[tex]V\mleft(cube\mright)=(15ft)^3=3375ft^3[/tex]

Adding the two volumes:

1125ft^3+3375ft^3=4500 cubic feet

In general, the y-intercept of the function F(x) = a • bx is the point _____.A.(0, b)B.(0, a)C.(0, x)D.(0, 1)

Answers

The y-intercept of a function is the point where the function crosses the y axis and where x = 0

[tex]\begin{gathered} We\text{ are asked to find the y intercept of an exponential function, y = a*b}^x \\ When\text{ x = 0, b}^x\text{ =1 for any value of b} \\ We\text{ are then left with y = a*1 when x =0} \end{gathered}[/tex]

The y intercept is therefore given by:

(0,a) --> option B

The senior classes at High School A and High School B planned separate trips to New York City.The senior class at High School A rented and filled 2 vans and 6 buses with 366 students. HighSchool B rented and filled 6 vans and 3 buses with 213 students. Each van and each bus carriedthe same number of students. Find the number of students in each van and in each bus.Answer: A van hasstudents and bus has students

Answers

Let V be the number of students that fit inside a van and B the number of students that fit inside a bus. Since 366 students fit in 2 vans and 6 buses, then:

[tex]2V+6B=366[/tex]

Since 213 students fit in 6 vans and 3 buses, then:

[tex]6V+3B=213[/tex]

Multiply the second equation by 2:

[tex]\begin{gathered} 2(6V+3B)=2(213) \\ \Rightarrow12V+6B=426 \end{gathered}[/tex]

Then, we have the system:

[tex]\begin{gathered} 2V+6B=366 \\ 12V+6B=426 \end{gathered}[/tex]

Substract the first equation from the second one and solve for V:

[tex]\begin{gathered} (12V+6B)-(2V+6B)=426-366 \\ \Rightarrow12V-2V+6B-6B=60 \\ \Rightarrow10V=60 \\ \Rightarrow V=\frac{60}{10} \\ \therefore V=6 \end{gathered}[/tex]

Substitute V=6 into the first equation and solve for B:

[tex]\begin{gathered} 2V+6B=366 \\ \Rightarrow2(6)+6B=366 \\ \Rightarrow12+6B=366 \\ \Rightarrow6B=366-12 \\ \Rightarrow6B=354 \\ \Rightarrow B=\frac{354}{6} \\ \therefore B=59 \end{gathered}[/tex]

Therefore, a van has 6 students and a bus has 59 students.

Question 2 - Minimum Hours f. In the previous question, if Leah babysits for 7 hours this month, what is the minimum number of hours she would have to work at the ice cream shop to earn at least $120? Justify your answer. (4 POINTS) Give your answer to the nearest whole hour.

Answers

Leah earns 5x + 8y dollars, after x hours babysitting and y hours at the ice cream shop.

She wants to earn at least $120, then:

5x + 8y ≥ 120

Given that Leah babysits for 7 hours, then:

5(7) + 8y ≥ 120

35 + 8y ≥ 120

8y ≥ 120 - 35

8y ≥ 85

y ≥ 85/8

y ≥ 10.625

She must work at least 11 hours

Given that 7 + 11 = 18, then she would not work more than 20 hours as she expected

Let Iql = 5 at an angle of 45° and [r= 16 at an angle of 300°. What is 19-r|?13.00 14.2O 15.518.0

Answers

As given that:

[tex]|q|=5[/tex]

At angle of45 degree

and |r| = 16 at 300 degree

so the |q| at 300 degree is:

[tex]\begin{gathered} |q|=5\times\frac{300}{45} \\ |q|=33.33 \end{gathered}[/tex]

Now |q-r| is:

[tex]\begin{gathered} |q-r|=33.33-16 \\ |q-r|=17.33 \\ |q-r|\approx18 \end{gathered}[/tex]

So the correct option is d.

having a bit of a problem with this logarithmic equation I will upload a photo

Answers

SOLUTION

We are asked to solve the equation

[tex]4^{5x-6}=44[/tex]

44 cannot be written in index form. So to solve this, we will take logarithm of both sides of the equation

We will have

[tex]\log 4^{5x-6}=\log 44[/tex]

Solving for x, we have

[tex]\begin{gathered} \log 4^{5x-6}=\log 44 \\ \\ 5x-6\log 4=\log 44 \\ \\ \text{dividing both sides by log4} \\ \\ 5x-6=\frac{\log 44}{\log 4} \\ \\ 5x=\frac{\log44}{\log4}+6 \end{gathered}[/tex]

The exact solution becomes

[tex]x=\frac{(\frac{\log44}{\log4}+6)}{5}[/tex]

The approximate solution to 4 d.p

[tex]\begin{gathered} x=\frac{(\frac{\log44}{\log4}+6)}{5} \\ \\ x=\frac{(\frac{1.64345}{0.60206}+6)}{5} \\ \\ x=\frac{8.72971}{5} \\ \\ x=1.7459 \end{gathered}[/tex]

The ratio between the radius of the base and the height of a cylinder is 2:3. If it's volume is 1617cm^3, find the total surface area of the cylinder.

Answers

Solution:

The ratio of the radius to the height of the cylinder is

[tex]2\colon3[/tex]

Let the radius be

[tex]r=2x[/tex]

Let the height be

[tex]h=3x[/tex]

The volume of the cylinder is given below as

[tex]V=1617cm^3[/tex]

Concept:

The volume of a cylinder is given below as

[tex]V_{\text{cylinder}}=\pi\times r^2\times h[/tex]

By substituting values, we will have

[tex]\begin{gathered} V_{\text{cylinder}}=\pi\times r^2\times h \\ 1617=\frac{22}{7}\times(2x)^2\times(3x) \\ 1617=\frac{22}{7}\times4x^2\times3x \\ 1617\times7=264x^3 \\ \text{divdie both sides by 264} \\ \frac{264x^3}{264}=\frac{1617\times7}{264} \\ x^3=\frac{343}{8} \\ x=\sqrt[3]{\frac{343}{8}} \\ x=\frac{7}{2} \end{gathered}[/tex]

The radius therefore will be

[tex]\begin{gathered} r=2x=2\times\frac{7}{2} \\ r=7cm \end{gathered}[/tex]

The height of the cylinder will be

[tex]\begin{gathered} h=3x=3\times\frac{7}{2} \\ h=\frac{21}{2}cm \end{gathered}[/tex]

The formula for the total surface area of a cylinder is given below as

[tex]T\mathrm{}S\mathrm{}A=2\pi r(r+h)[/tex]

By substituting the values, we will have

[tex]\begin{gathered} TSA=2\pi r(r+h) \\ TSA=2\times\frac{22}{7}\times7(7+\frac{21}{2}) \\ TSA=44(7+\frac{21}{2}) \\ TSA=44\times7+44\times\frac{21}{2} \\ TSA=308+462 \\ TSA=770cm^2 \end{gathered}[/tex]

Hence,

The total surface area of the cylinder is = 770cm²

Each expression below represents the area of a rectangle written as a product tha area model for each expression on your paper and label its length and then we nation showing that the area written as a product is equal to the area wimen as the the parts de prepared to share your equations with the class. a. (x+3)(2x + 1) ca(224) d. (2x + 5)(x + y + 2) • (20 - 1999 (2x-1) (2x-1) g. 2(325) ( 2+ y + 3) 2

Answers

We have the expression:

[tex]x(2x-y)[/tex]

It can be thougth as the area of a rectangle with sides x and (2x-y).

We can also think of the the difference between an area of a rectangle with sides x and 2x and a rectangle with sides x and y:

[tex]x(2x-y)=x\cdot2x-x\cdot y=2x^2-xy[/tex]

Answer: x(2x-y) = 2x^2-xy

It took Carmen 4 hours to drive 200 miles. Using context clues from the problem, what formula can be used to find George's rate of speed?

Answers

From the information given, we can find th rate of speed with the equation;

[tex]\text{distance = rate x time}[/tex]

Or;

[tex]d=rt[/tex]

Does the point (–48, –47) satisfy the equation y = x − 1?

Answers

To find the answer to the question, we will substitute "-48" into "x" and "-47" into "y" and see if the equation holds true or not.

[tex]\begin{gathered} y=x-1 \\ -47\stackrel{?}{=}-48-1 \\ -47\neq-49 \end{gathered}[/tex]

Thus, the point (-48, -47) does not satisfy the equation y = x - 1.

AnswerNo

write and slove six less then the product of a number n and 1/4 is no more than 96 fill in the boxs

Answers

ANSWER:

[tex]n\leq408[/tex]

STEP-BY-STEP EXPLANATION:

With the statement we deduce the following inequality

[tex]\frac{1}{4}\cdot n-6\leq96[/tex]

Solving for n

[tex]\begin{gathered} 4\cdot\frac{1}{4}\cdot n-4\cdot6\leq4\cdot96 \\ n-24\leq384 \\ n\leq384+24 \\ n\leq408 \end{gathered}[/tex]

how many minutes until the heart beats 200 times

Answers

From the given table, we can read that the 200 beats is associated with the entry: "5 minutes", so that is the answer we pick.

which agrees with the first option in the provided list of possible answers.

Find the exact area of la circle if its circumference is 367 cm.

Answers

Given the circumference to be 367 cm.

Recall that the formula for the circumference of a circle is given as;

[tex]\begin{gathered} C=2\pi r \\ \Rightarrow367=2\pi r \end{gathered}[/tex]

If we make r the subject of the formula,

[tex]r=\frac{367}{2\pi}[/tex]

The area of a circle is given as;

[tex]\begin{gathered} A=\pi r^2 \\ \Rightarrow\pi(\frac{367}{2\pi})^2=\frac{\pi}{4\pi^2}(367)^2=10718.21 \end{gathered}[/tex]

what is 6q - q please

Answers

To solve this expression, we just have to subtract because they are like terms

[tex]6q-q=5q[/tex]

Hence, the answer is 5q.

I need the answer to number 2 please answer it like the paper so that I can understand it better. Please

Answers

Elijah made an error of subtracting the numbers and not including the imaginary sign (letter i)

The correct midpoint is (6, 3i)

Explanation:

The two points are 8 + 4i and 4 + 2i

Elijah got the midpoint as (2, 1).

To determine Elijah's error, let's calculate the midpoint of a complex number:

[tex]\text{Midpoint = (}\frac{a\text{ + b}}{2}),\text{ (}\frac{c\text{ + d}}{2})i[/tex]

let 8 + 4i = a + ci

let 4 + 2i = b + di

The real numbers will be added together. The imaginary numbers will also be added together.

substituting the values in the formula:

[tex]\begin{gathered} \text{Midpoint = (}\frac{8\text{ + 4}}{2}),\text{ (}\frac{4\text{ + 2}}{2})i \\ \text{Midpoint = }\frac{12}{2},\text{ }\frac{6}{2}i \\ \\ \text{Midpoint = (6, 3i)} \end{gathered}[/tex]

Elijah made an error of subtracting the numbers. Also Elijah didn't include the letter representing the imaginary numbers (i).

The correct midpoint is (6, 3i)

For each value of w, determine whether it is a solution to w < 9.Is it a solution?W5?YesNo75914

Answers

Answer: 5 and 7

Explanation:

we need to determine if a number is a solution to

[tex]w<9[/tex]

That reads as "w is less than 9, and not equal to 9"

so we find in our options wich ones are less than 9. The options are:

• 7

,

• 5

,

• 9

,

• 14

The ones smaller or less than 9 are: 5 and 7

The ones greater than 9 or equal to 9 (the ones that are not the solution) are: 9 and 14.

So the solutions are: 5 and 7

Drag and drop the expressions into the boxes to correctly complete the proof of the polynomial identity.(x2 + y2)2 + 2x?y– y4 = x(x² + 4y?)(x2 + y2)2 + 2x²y2 – y4 = x(+ 4y?)+2x²y2 – y4 = x2 (x2 + 4y?)x² (x² + 47²)= x2 (x2 + 4y2)x² (x² + 47²) x² – 2x²y² + y x² + yt x² + 4x²72 x + 2x²,2x² + 2x²y² + yt

Answers

Answer:

x^4 + y^4 + 2x^2 y^2

x^4 + 4x^2y^2

x^2 (x^2 + 4y^2 )

Explanation:

Expanding the the expression gives

[tex]\begin{gathered} (x^2+y^2)^4=(x^2)^2+(y^2)^2+2(x^2)(y^2) \\ =\boxed{x^4+y^4+2x^2y^2\text{.}} \end{gathered}[/tex]

Simplifying the Left-hand side gives

[tex]\begin{gathered} x^4+y^4+2x^2y^2+2x^2y^2-y^4 \\ =\boxed{x^4+4x^2y^2\text{.}} \end{gathered}[/tex]

Finally, factoring out x^2 from the left-hand side gives

[tex]x^4+4x^2y^2=\boxed{x^2\mleft(x^2+4y^2\mright)\text{.}}[/tex]

laws of exponent : multiplication  and power to a poweranswer and help me step by step

Answers

[tex]\begin{gathered} 2(6c^2d^4e^5)^2=2(6)^2c^{2\times2}d^{4\times2}e^{5\times2} \\ =2(36)c^4d^8e^{10} \\ =72c^4d^8e^{10} \end{gathered}[/tex]

All the numbers multiply in the normal way, and the powers of a power need to be multiplied.

[tex]72c^4d^8e^{10}[/tex]

For which equation is x = 5 a solution ?

Answers

Given x = 5

We will find which equation will give a solution x = 5

1) x/2 = 10

so, x = 2 * 10 = 20

So, option (1) is wrong

2) x - 7 = 12

x = 12 + 7 = 19

So, option 2 is wrong

3) 2 + x = 3

x = 3 - 2 = 1

So, option 3 is wrong

4) 3x = 15

x = 15/3 = 5

So, the answer is option 3x = 15

Find the z-score location of a vertical line that separates anormal distribution as described in each of the following.a. 15% in the tail on the rightb. 40% in the tail on the leftc. 75% in the body on the rightd. 60% in the body on the left

Answers

Answer:

a. z = 1.0364

b. z = -0.2533

c. z = -0.6745

d. z = 0.2533

Explanation:

We can represent each option with the following diagrams

So, for each option, we need to find a z that satisfies the following

a. P(Z > z) = 0.15

b. P(Z < z) = 0.40

c. P(Z > z) = 0.75

d. P(Z > z) = 0.60

Then, using a normal table distribution, we get that each value of z is

a. z = 1.0364

b. z = -0.2533

c. z = -0.6745

d. z = 0.2533

In 6-13 round each number to the place of the underlined digit

Answers

6. 32.7

7. 3.25

8. 41.1

9. 0.41

10. 6.1

11. 6.1

12. 184

13. 905.26

1) Considering that the underline marks the place to be rounded off we can do the following:

Note that if the number is greater than or equal to 5 then we will round it up.

If the number is lesser than 5 it will be rounded down.

Based on that we can round like this.

6. 32.7

7. 3.25

8. 41.1

9. 0.41

10. 6.1

11. 6.1

12. 184

13. 905.26

2)

Find the roots of the equation 5x2 + 125 = 0

Answers

Answer:[tex]5i\text{ and -5i}[/tex]Explanation:

The given equation is:

[tex]5x^2+125=0[/tex]

Divide through by 5

[tex]\begin{gathered} \frac{5x^2}{5}+\frac{125}{5}=\frac{0}{5} \\ \\ x^2+25=0 \\ \end{gathered}[/tex]

This is further simplified as:

[tex]\begin{gathered} x^2=-25 \\ \\ \sqrt{x^2}=\pm\sqrt{-25} \\ \\ \sqrt{x^2}=\pm\sqrt{-1}\times\sqrt{25} \\ \\ x^=\pm5i \\ \\ x=5i\text{ and -5i} \\ \end{gathered}[/tex]

Solve for x: |x - 2| + 10 = 12 A x = 0 and x = 4B x = -4 and x = 0C x = -20 and x=4 D No solution

Answers

|x - 2| + 10 = 12

|x - 2| = 12 -10

|x - 2| = 2

There are 2 solutions:

x-2 = 2 and x-2 = -2

Solve each:

x = 2+2

x = 4

x-2=-2

x =-2+2

x=0

solution: x=0 and x = 4

Go step by step to reduce the radical. V 243 DVD try You must answer all questions above in order to submit.

Answers

We are given the following radical

[tex]\sqrt[]{243}[/tex]

Let us reduce the above radical

We need to break the number 243 into a product of factors

Notice that 81 and 3 are the factors (83×3 = 243)

[tex]\sqrt[]{243}=\sqrt[]{81}\cdot\sqrt[]{3}[/tex]

Since 81 is a perfect square so the radical becomes

[tex]\sqrt[]{243}=\sqrt[]{81}\cdot\sqrt[]{3}=9\cdot\sqrt[]{3}[/tex]

Therefore, the simplified radical is

[tex]undefined[/tex]

The area of Square A is 36 square cm. The area of Square A’(A Prime) is 225 ᶜᵐ². What possible transformations did the square undergo? 

Answers

A possible transformation is a scale. Since the area changed by

[tex]\frac{225}{36}=\frac{25}{4}[/tex]

then a possible transformation was a scale by 25/4. A scale by a ratio bigger than one is a dilation.

Then the answer is B.

Other Questions
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