If we use the distributive property on the second expression, we get the following:
[tex]-4\cdot(3a-1)=-4\cdot(3a)-4(-1)=-12a+4[/tex]therefore, the expressions are not equivalent
find the slope of the line through the points (-6,5) and (3, -2)
We have to find the slope of the line that pass through points P1=(-6,5) and P2=(3,-2).
We can calculate it as:
[tex]m=\frac{y_2-y_1}{x_2-x_1}=\frac{-2-5}{3-(-6)}=\frac{-7}{9}=-\frac{7}{9}[/tex]Answer: the slope of the line is m = -7/9
Can you help me figure out how to find the original radican ??? I have no clue how to do so
So we have:
[tex]-3a^5b^2\sqrt[3]{a^2c}[/tex]And we want to knowthe original before simplification, that is, before evaluating the interior part of the root.
So, we need to figure a way to put the part outside of the root back in.
Taking the cubic root of a number is the same as dividing its exponent by 3, because:
[tex]\sqrt[3]{a^n}=a^{\frac{n}{3}}[/tex]So, thinking in the other direction, we need to multiply the exponents by 3 before taking it back to the inside of the cubic root:
[tex]a^k=a^{\frac{3k}{3}}=\sqrt[3]{a^{3k}}[/tex]So, the b part have a 2 in the exponent, so we can multiply it by 3 to get 6:
[tex]\begin{gathered} b^2=b^{\frac{3\cdot2}{3}}=\sqrt[3]{b^{3\cdot2}}=\sqrt[3]{b^6} \\ -3a^5b^2\sqrt[3]{a^2c}=-3a^5\sqrt[3]{b^6}\sqrt[3]{a^2b^{3\cdot2}c}=-3a^5\sqrt[3]{a^2b^6c} \end{gathered}[/tex]The a part have a 5 in the exponent, so we will get 15:
[tex]\begin{gathered} a^5=a^{\frac{3\cdot5}{3}}=\sqrt[3]{a^{3\cdot5}}=\sqrt[3]{a^{15}} \\ -3a^5\sqrt[3]{a^2b^6c}=-3\sqrt[3]{a^{15}}\sqrt[3]{a^2^{}b^6c}=-3\sqrt[3]{a^2a^{15}b^6c} \end{gathered}[/tex]Now, since we have a² and a¹⁵, we can add their exponents:
[tex]\begin{gathered} a^2a^{15}=a^{17} \\ -3\sqrt[3]{a^2a^{15}b^6c}=-3^{}\sqrt[3]{a^{17}b^6c} \end{gathered}[/tex]Now, the -3 have an exponent of 1, so:
[tex]\begin{gathered} -3=(-3)^1=(-3)^{\frac{3\cdot1}{3}}=\sqrt[3]{(-3)^{3\cdot1}}=\sqrt[3]{(-3)^3}=\sqrt[3]{-27} \\ -3^{}\sqrt[3]{a^{17}b^6c}=\sqrt[3]{-27}^{}\sqrt[3]{a^{17}b^6c}=^{}\sqrt[3]{-27a^{17}b^6c} \end{gathered}[/tex]Thus, we have, in the end:
[tex]^{}\sqrt[3]{-27a^{17}b^6c}=-3a^5b^2^{}\sqrt[3]{a^2^{}c}[/tex]Write the equation of the circle:center at (5, - 2) , passes through (4, 0)
the equation of the circle is
(x-h)^2 + (y-k)^2 = r^2
if we replace the terms
(4-5)^2 + (0-(-2))^2 = r^2
Now, we can fin the radius if we solve the previous equation
( - 1 )^2 + ( 2 )^2 =r^2
1 + 4 = r^2
5 = r^2
r = SQRT(5)
Now, since we already know r, we can replace it in the circle equation to obtain the result
so, (x-h)^2 + (y-k)^2 = r^2
iquals to, (x-5)^2 + (y+2)^2 = 5
Find the perimeter of the shaded region of this composite figure .You can use 3.14 for pi.llAlso round the answer to the nearest hundreth.
ANSWER
18.58 m
EXPLANATION
We need to find the perimeter of the shaded region of the figure.
The figure is made up of a rectangle with the cut out of a semi-circle, so, to find the perimeter, we will subtract the perimeter of the semi-circle (without the diameter) from that of the rectangle.
The perimeter of the rectangle is:
P = 2(L + B)
where L = length = 8 m
B = breadth = 6 m
So, the perimeter of the rectangle is:
P = 2(8 + 6) = 2 * 14
P = 28 m
The perimeter (circumference) of the semi-circle (without the diameter) is:
C = π * R
where R = radius of the semicircle
The diameter is 6 m, so the radius is:
R = D / 2 = 6 / 2 = 3 m
So, the circumference of the semicircle is:
C = 3.14 * 3
C = 9.42 m
So, the perimeter of the composite figure is:
P = 28 - 9.42
P = 18.58 m
That is the answer.
Find the 5and term of the arithmetic sequence 5, 9, 13,
Notice that:
[tex]\begin{gathered} 9-5=4, \\ 13-9=4. \end{gathered}[/tex]Since the sequence is arithmetic then, the nth term has the following form:
[tex]a_n=5+4\cdot(n-1)\text{.}[/tex]Therefore:
[tex]a_{52}=5+4(51)=5+204=209.[/tex]Answer: 209.
what is the quotient of the complex numbers below 3 + 2i / 1 - 5i
Take the conjugate of the denominator, use it to multiply the numerator and the denominator
That is;
[tex]undefined[/tex]Answer:
Step-by-step explanation:
Lakshmi bought 7 books for a total of 56 rupees how much would see pay for just three books? 56 rupees Indian money
To find how much would be paid for 3 books, follow the steps below.
Step 01: Find the price of one book.
Let's say the price of one book is x.
Then, the price of 7 books is 7 times x, which is 56 rupes.
[tex]7x=56[/tex]To find x, let's divide both sides by 7:
[tex]\begin{gathered} \frac{7x}{7}=\frac{56}{7} \\ 1x=8 \\ x=8 \end{gathered}[/tex]So, the price of one book is 8.
Step 02: Find the price of 3 books.
If the price of one book is 8, the price of 3 books (P) will be 3 times 8:
[tex]\begin{gathered} P=3\cdot8 \\ P=24 \end{gathered}[/tex]Answer: It would be paid 24 rupees for 3 books.
Last year, Emma went bowling several times and earned an average score of 130 points. This
year, after taking a class at school, she improved her score to an average of 234 points. What
is the percent of increase in Emma's average score?
Answer:
80%
Step-by-step explanation:
234-130
=104
(n×p)×100=answer
(130×p)÷100= 10
(130 × p) ÷ 100=104
(130 × p) ÷ 100) × 100 = 104 × 100
130p = 10400
130p ÷ 130 = 10400 ÷ 130
p = 80%
1. Consider the following functions. f(x) = 3x2 + x + 2 g(x) = 4x2 + 2(3x – 4) h(x) = 5(x2 - 1) a. Find f (x) - g(x). b. Find g(x) - h(x).
a.
Let's write function g(x) better:
[tex]g(x)=4x^2+2(3x-4)=4x^2+6x-8[/tex]Now we can do the substraction easier
[tex]f(x)-g(x)=(3x^2+x+2)-(4x^2+6x-8)_{}[/tex][tex]f(x)-g(x)=3x^2+x+2-4x^2-6x+8[/tex][tex]f(x)-g(x)=(3-4)x^2+(1-6)x+2^{}+8[/tex][tex]f\mleft(x\mright)-g\mleft(x\mright)=-x^2-5x+10[/tex]That's answer a
b.
We write h(x) better too:
[tex]h(x)=5(x^2-1)=5x^2-5[/tex]And do the same as before:
[tex]g(x)-h(x)=(4x^2+6x-8)-(5x^2-5)[/tex][tex]g(x)-h(x)=4x^2+6x-8-5x^2+5[/tex][tex]g(x)-h(x)=(4-5)x^2+6x-8+5[/tex][tex]g(x)-h(x)=-x^2+6x-3[/tex]That's answer b
Yovanni went on a hike. He climbed 4/5 of mile in 1/4 of an hour. What was his hiking speed in miles per hour
We have the following information
Distance
[tex]d=\frac{4}{5}\text{miles}[/tex]Time
[tex]t=\frac{1}{4}\text{hours}[/tex]To find his hiking speed we need to use the formula for speed:
[tex]s=\frac{d}{t}[/tex]where d is the distance and t is the time.
We substitute our values into the formula:
[tex]s=\frac{\frac{4}{5}\text{miles}}{\frac{1}{4}\text{hours}}[/tex]In this type of divisions, we multiply the extremes of the expression (4 by 4) and this will be our numerator. Also, we multiply the numbers in the middle (5 by 1) and this will be our denominator:
[tex]s=\frac{4\times4}{5\times1}=\frac{16}{5}=3.2\text{ mi/h}[/tex]Answer: 3.2 mi/h
Calculate the maximum number of cylindrical paint cans that carvers auto custom can stock if the paint comes in a 2-pack hazmat box that mesures 15 inches by 7 inches by 6 inches
The volume of Hazmat box is,
[tex]v=15\times7\times6[/tex][tex]v=630in^3[/tex]Convert inches to feet ,
[tex]undefined[/tex]The volume of the warehouse when half of the warehouse is painted with cams and rims is.
[tex]V=\frac{8000}{2}\times20ft^3[/tex][tex]V=80,000ft^3[/tex]Suppose f(x) = x². Find the graph off(x+3).???
If f(x)=x^2
Then f(x+3)=(x+3)^2
[tex](x+3)^2=x^2+6x+9[/tex]Use geogebra to graph the function or calculate the vertex using the equation
x=-b/2a
From the equation we have that
b=6
a=1
x=-6/(2*1)
x=-3
The vertex is on x=-3
Calculate f(-3)=9-18+9=0
The vertex is (-3,0)
y axis cut off point f(0)=0+0+9=9
As "a" is a positive value, parabola open upwards, now you can draw the parabola
This is a sketch, let's use geogebra
Solve the inequality algebraically. Express your answer using set notation or interval notation. l x-8l greater than or equal to 4. Rewrite the inequality without the absolute values.
To rewrite the inequality:
[tex]\lvert{x-8}\rvert\ge4[/tex]we need to remember that:
[tex]\lvert{x}\rvert\ge a\text{ is equivalent to }x\ge a\text{ or }x\leq-a[/tex]Then in this case we have:
[tex]\begin{gathered} \lvert{x-8}\rvert\ge4 \\ \text{ Is equivalent to:} \\ x-8\ge4\text{ or }x-8\leq-4 \end{gathered}[/tex]Therefore, we can rewrite the inequality as:
[tex]x-8\leq-4\text{ or }x-8\ge4[/tex]Once we have it written in this form we can solve it:
[tex]\begin{gathered} x-8\leq-4\text{ or }x-8\ge4 \\ x\leq-4+8\text{ or }x\ge4+8 \\ x\leq4\text{ or }x\ge12 \end{gathered}[/tex]Therefore, the solution set of the inequality is:
[tex](-\infty,4\rbrack\cup\lbrack12,\infty)[/tex]HELP. i am so confused. the question is in the picture
1) If we consider that y=f(2x) is a transformed version of y=f(x) then we can set a t-table and plug for the given point (16,8) the x-coordinate x=16
[tex]\begin{gathered} (16,8)-\longrightarrow f(x)-->y=\frac{1}{2}x \\ \end{gathered}[/tex]Since the new function, f(2x) requires us to divide the input by 2 to compensate
Solve 431 ÷ 3 on paper. You'll see that there is a remainder.
What digit in the ones place would give us no remainder?
Answer:
2 in the ones place.
Step-by-step explanation:
431/3 won't work because 4+3+1=NOT multiple of 3.
The closest number in the ones place that will make it an integer is 2, because 4+3+2=multiple of 3
Answer:
2, 5, or 8.
Step-by-step explanation:
There is actually a trick to this one.
---
If the digits in a number add up to a multiple of 3, then the whole number is divisible by 3.
For example:
843
Add the digits:
8+4+3=15
You could add the digits again:
1+5=6
Six is a multiple of 3, so 843 is a multiple of 3.
---
Now your number was 431.
4+3+1=8
8 is not divisible by 3, so 431 is not divisible by 3.
432, 435, and 438 would work in this situation.
PLEASE MARK AS BRANLIEST!!Why is x² + 36 NOT factorable? In other words, why is it prime? What are twodetails that draw you to this conclusion?
SOLUTION:
Step 1:
In this question, we are given the following:
a) Why is x² + 36 NOT factorable?
b) In other words, why is it prime?
c) What are two details that draw you to this conclusion?
Step 2:
The details of the solution are as follows:
[tex]\begin{gathered} a)\text{ x}^2\text{ + 36 is not factorizable under of field of integers Z,} \\ since\text{ it cannot be expressed as product of two squares} \end{gathered}[/tex]b) In other words, why is it prime?
It is a prime polynomial because a prime polynomial is one that cannot be factored into the product of two polynomials, using integer values.
-
c) What are two details that draw you to this conclusion?
1) You can factor a difference of squares, but not a sum of squares.
2) A prime polynomial is one that cannot be factored into the product of two polynomials, using integer values.
-
A rocket is launched by Team Flash from the ground on Earth-73. The rocket passes a sensor at a height of5760 feet after 8 seconds and lands back on Earth-73 after 53 seconds.Write an equation for the height of the rocket, h, in feet as a function of the number of seconds, t, since therocket was launched.Round to 3 decimal places as needed.After how many seconds will the rocket reach its maximum height?Round to 3 decimal places as needed.What is the maximum height in feet that the rocket reaches?Round to 3 decimal places as needed.
We know two points of the trajectory of the rocket:
1) A height of 5760 ft at time t=8 seconds after launch.
2) A height of 0 ft (landing) at time t=53 seconds after launch.
We also know that the initial position was a height of 0 ft at t=0 seconds.
So we have 3 points to write the equation, that will be a quadratic equation for this kind of trayectory.
As we know we have roots at t=0 and t=53, we can start writing it as:
[tex]h(t)=a(t-0)(t-53)=at(t-53)[/tex]We have one point left, (t,h) = (8, 5760), to find the parameter "a". We can replace t and y in the equation and solve as:
[tex]\begin{gathered} h(t)=at(t-53) \\ 5760=a\cdot8\cdot(8-53) \\ 5760=a\cdot8\cdot(-45) \\ 5760=a\cdot(-360) \\ a=\frac{5760}{-360} \\ a=-16 \end{gathered}[/tex]Then we can write the equation as:
[tex]h=-16t(t-53)=-16t^2+848t[/tex]We can graph it as:
In this kind of trajectories, the maximum height is reached halfway between the launch and the landing.
For any function, we can find the maximum of minimums deriving the function and equal it to 0. We will do it for this function:
[tex]\begin{gathered} \frac{dh}{dt}=-16(2t)+848=0 \\ -32t+848=0 \\ 32t=848 \\ t=\frac{848}{32} \\ t=26.5 \end{gathered}[/tex]The maximum height is reached at time t=26.5 seconds.
Now we can calculate the height at t=26.5 seconds, the maximum height, as:
[tex]\begin{gathered} h(26.5)=-16(26.5)^2+848(26.5) \\ h(26.5)=-16\cdot702.25+22472 \\ h(26.5)=-11236+22472 \\ h(26.5)=11236 \end{gathered}[/tex]Answer:
a) The equation is h(t) = -16t²+848t
b) The maximum height is reached at time t=26.5 seconds.
c) The maximum height is 11236 ft.
I NEED HELP QUICKLY ITS DUE 8PM AND I HAVE OTHER HOMEWORK TO DO.
Answer:
$5625
Explanation:
The equation for your earning y = 150x - x² is the equation of a parabola, so the maximum point of the parabola has a coordinate x equal to -b/2a
Where b is the number beside the x and a is the number beside the x²
In this case, a is -1 and b is 150, so the x-coordinate of the maximum is:
[tex]x=\frac{-b}{2a}=\frac{-150}{2(-1)}=\frac{-150}{-2}=75[/tex]With the value of x, we can calculate the value of y, so:
y = 150x - x²
y = 150(75) - (75)²
y = 11250 - 5625
y = 5625
Therefore, the maximum amount that you can earn is $5625.
Solve for 3x/2 -4 = 16what does x equal??
The given equation is expressed as
3x/2 - 4 = 16
The first step is to multiply both sides of the equation by 2. It becomes
3x/2 * 2 - 4 * 2 = 16 * 2
3x - 8 = 32
3x = 32 + 8
3x = 40
x = 40/3
x = 13.33
There is a population of 405,000 bacteria in a colony. If the number of bacteria doubles every 44 hours, what will the population be 176 hours from now?
Since the population doubles every 44 hours, it can be modeled using an exponential equation as follows:
[tex]P(t)=405,000\times2^{\frac{t}{44}}[/tex]Where t is the time since the population was 405,000 measured in hours.
Replace t=176 to find the population after 176 hours:
[tex]\begin{gathered} P(176)=405,000\times2^{\frac{176}{44}} \\ =405,000\times2^4 \\ =405,000\times16 \\ =6,480,000 \end{gathered}[/tex]Therefore, the population after 176 hours will be 6,480,000
Write a coordinate proof of the following theorem:"If a quadrilateral is a kite, then its diagonals are perpendicular."(image attached)thank you ! :)
Given the following coordinate points from the kite
[tex]\begin{gathered} W=(a,4b) \\ X=(2a,b) \\ Y=(a,0) \\ Z=(0,b) \end{gathered}[/tex]For the diagonals to be perpendicular the product of the distance WY and XZ must be zero that is;
[tex]\vec{WY}\cdot\vec{XZ}=0[/tex]Determine the coordinate point WY
[tex]\begin{gathered} \vec{WY}=[(a-a,4b-0)] \\ \vec{WY}=(0,4b) \end{gathered}[/tex]Determine the coordinate point XZ
[tex]\begin{gathered} \vec{XZ}=[(2a-0),(b-b)] \\ \vec{XZ}=(2a,0) \end{gathered}[/tex]Take the dot product of the coordinates
[tex]\begin{gathered} \vec{WY}\cdot\vec{XZ}=(0,4b)\cdot(2a,0) \\ \vec{WY}\cdot\vec{XZ}=[(0)(2a)+(4b)(0))] \\ \vec{WY}\cdot\vec{XZ}=(0,0)=\vec{0} \end{gathered}[/tex]Since the dot product of the coordinates is a zero vector, hence its diagonals are perpendicular.
WILL MARK BRAINLIEST Rectangle PQRS is shown above. Point C is the center of the rectangle.Maggle claims that there are transformations that preserve the length of the rectangle's sides. Which of the following transformations could be used to support Maggie's claim? select all that apply1.) a translation of 10 units to the right2.) a rotation of 90' clockwise about vortex Q3.) a reflection over the side RS4.) a diation of scale factor 1 through contor5.) a vertical stretch of scale factor 2 through contor C
With the given options;
(1) A translation of 10 units to the right will only map the rectangle onto a different location, but the lengths would remain as it were
(3) A reflection over the side RS will turn the rectangle into a mirror reflection of itself, and this means the sides remain the same measurement but is now being observed from the opposite side (but top now bwcomes bottom and vice versa).
(4) A dilation of scale factor 1 through center
Solve the equation 42+7c - 5 = 0 using the quadratic formula
The equation is:
[tex]4c^2+7c-5=0[/tex]so we can use the cuadratic equation so:
[tex]\begin{gathered} c=\frac{-b\pm\sqrt[]{b^2-4(a)(c^{\prime})}}{2(a)} \\ \text{where} \\ a=4 \\ b=7 \\ c^{\prime}=-5 \end{gathered}[/tex]So if we replace in the equation we will have:
[tex]c=\frac{-7\pm\sqrt[]{7^2-4(4)(-5)}}{2(4)}[/tex]So we simplify to solve the problem so:
[tex]c=\frac{-7\pm\sqrt[]{129}}{8}[/tex]My dog got hurt and needed surgery,so I had to use my credit card to pay the vet bill. His surgery costed me $5,323.21. If my monthly interest rate is 1.42%, how much is my finance charge for the first billing cycle?
Answer:
$75.59.
Explanation:
• Cost of the surgery = $5,323.21
,• Monthly interest rate = 1.42%
A finance charge is a fee charged for the use of a credit card. A billing cycle is usually between 28 to 31 days, i.e. a month.
To find the finance charge, multiply the interest rate by the cost of surgery.
[tex]\begin{gathered} \text{Fnance Charge}=1.42\%\text{ of \$}5,323.21 \\ =\frac{1.42}{100}\times5,323.21 \\ =\$75.59 \end{gathered}[/tex]The finance charge for the first billing cycle is $75.59.
determine whether the equation below has a one solutions, no solutions, or an infinite number of solutions. afterwards, determine two values of x that support your conclusion. 4(x+4) = 4x+16the equation has ____ solutions.a value of x that makes the equation true is __,which when simplified makes the equation turn into____=_____.a value of x that makes the equation false is____, which when simplified makes the equation turn into ___=___.
4(x+4) = 4x+16
Apply distributive property:
4(x)+4(4)= 4x+16
4x+16=4x+16
Add and subtract alike terms
4x-4x= 16-16
0=0
Since x can have many values, it has an infinite number of solutions.
we can replace x by 1, by 3, by 2 and the equality will remain.
the equation has an infinite number of solutions.
A student finished 45 of her homework problems in class. If the ratio of problems shefinished to problems she still had left was 9:4, how many homework problems did shehave total?
The ratio between the amount of problems she finished and the problems she still had left is given by the division between those amount. Since this ratio is 9:4, if we call the amount of homework she still has left as x, we have the following relation
[tex]\frac{45}{x}=\frac{9}{4}[/tex]Solving for x, we have
[tex]\begin{gathered} \frac{45}{x}=\frac{9}{4} \\ \frac{x}{45}=\frac{4}{9} \\ x=\frac{4}{9}\cdot45 \\ x=\frac{4\cdot45}{9} \\ x=\frac{180}{9} \\ x=20 \end{gathered}[/tex]She still has 20 problems left to solve. The total amount of problems is given by the sum between the problems she already finished and the problems left to solve, then, the total amount of problems is
[tex]45+20=65[/tex]65 problems.
If f(x) = -x² - 2x, what is f(-2)?
Answer: 0
Step-by-step explanation:
f(-2)= -(-2)^2-2(-2)
= -4+4=0
Factor the polynomial completely.X^2+x+1
1) Examining the expression below, we can group the first and the second term:
[tex]\begin{gathered} x^2+x+1 \\ x(x+1)+1 \\ \end{gathered}[/tex]Note that there is no way beyond this point. So we could not factor beyond this point.
2. Which answer of
the following is an
example of a SUM?
A 12-3=9
B 12+3=15
C 12×3=36
D 12÷3=4
Answer: B : 12+3=15
Step-by-step explanation:
A sum is the answer of an addition problem
There are two buildings beside a park.
The first building is 165 3/4 ft tall, and
the second building is 114 1/4 ft tall.
By rounding to the nearest whole number,
estimate the difference between the
heights of the buildings.
The difference of the height of the buildings is 51.5 for the first building is 165 3/4 feet tall, and the second building is 114 1/4 feet tall.
Given that,
There are two building side of a park.
The first building is 165 3/4 feet tall, and the second building is 114 1/4 feet tall.
We have to find the difference of the height of the buildings.
We have to subtract the first building is 165 3/4 feet tall, and the second building is 114 1/4 feet tall.
165 3/4- 114 1/4
660+3/4- 456+1/4
663/4-457/4
165.75-114.25
51.5
Therefore, The difference of the height of the buildings is 51.5 for the first building is 165 3/4 feet tall, and the second building is 114 1/4 feet tall.
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