Find the area of IJKL

Answer:

volume = 2 x 2 x 2 = 8 cm²

Step-by-step explanation:

Answer:

x=57

Step-by-step explanation:

a triangle has a sum of 180 degrees

that angle 51 is vertical to the angle across it so it is 51

we do math now

180-72-51=57

I hope this helped :))

Answer:

I think it's C

Step-by-step explanation:

Answer:

(4, 1)

Step-by-step explanation:

Since there's a graph, the answer is the intersection: (4, 1)

But here's the cancellation method for future use :)

Answer:

x = 23

y = 17

Step-by-step explanation:

x = 1st number

y = 2nd number

System of Equations:

x + y = 40

x - y = 6;    this can be used as x = 6+y

6+y + y = 40

6 + 2y = 40

2y = 34

y = 17

x + 17 = 40

x = 23

Answer:

The answer is

90 degrees the first one is I don't know about the rest sorry

Step-by-step explanation:

Answer:

No solution

Step-by-step explanation:

because the are parallel lines they never meet so they are inconsistent so there is no solution

Answer:

answer is B

Step-by-step explanation:

he goes up

Answer: There were 525 pine trees.

Step-by-step explanation:

Given: Total trees = 1250

Percentage of oak trees = 32%

Percentage of birch trees = 26%

Since remaining trees are pine trees.

Percentage of pine trees = 100% - 32% - 26%

=42%

The number of pine trees = 42% of 1250

= 0.42 ×1250

= 525

Hence, there were 525 pine trees.

Answer:

20 ≥ x

Step-by-step explanation:

Alex has $20 or less to spend at the store

9514 1404 393

Answer:

Step-by-step explanation:

The zeros are the values of t that make the factors zero.

  1 -t = 0   ⇒   t = 1

  8 +16t = 0   ⇒   t = -8/16 = -1/2

The equation is used to model height after the ball is thrown. We don't expect it to be a good model before the ball is thrown (t < 0), so the zero in that region is extraneous.

Only the positive zero is in the function's domain, so that is the only one that is meaningful.

__

When t = 0 (at the time the ball is thrown), the function value is ...

  h(0) = (1 -0)(8 +0) = 8

The ball is thrown from a height of 8 feet.

Answer:

because they are both similar. they both accept the law of karma and release from the cycle of birth. one of the key differences is

buddhist only believe in one god. hinduism believes in multiple gods

Answer:

hey emanii

Step-by-step explanation:

Answer:

Does this count as stander form 6.95880148018

Step-by-step explanation:

Answer:

y = 8 root 2

Step-by-step explanation:

Since it is a 90-45-45 triangle, y is s root 2

s = 8

so, y = 8 root 2

Answer:

y=8√2

Step-by-step explanation:

Because 2 of the angles are the same, the triangle is isosceles (because of base angles theorem). Therefore, x=8. Using Pythagoras theorem, y=8√2.

Answer:

chicken wing chicken wing hot dog and baloney chicken and macaroni chillen with my homie

Step-by-step explanation:

Answer:

Harry is paid £64

Step-by-step explanation:

To work this out you have to divide 48 ÷ 6 to get 8.

8 hours x £8 = £64

Answer:

64

you have to find the LCM of 48 including 6 and use the other number which is 8 and multiply it by how long Harry worked 8 hours which gives you £64

Answer:

34%

Step-by-step explanation:

I meant be a couple PREcents OFF BUT IM SORRY THIS IS MY BEST ANSWR :( hope this helps tho :)

Answers:

So,

The general rule is that the upper boundary is found computing Q3+1.5*IQR, where IQR is the interquartile range. Recall that IQR = Q3-Q1. The Q1 and Q3 are the first and third quartiles respectively.

Another rule is that the lower boundary is found by computing Q1 - 1.5*IQR

On a box and whisker plot, outliers are points that are not connected to the main plot itself. You can think of them as islands far from the mainland.

What are Outliers?

If a number in distribution is more than 1.5 times the length of the box away from either of the upper and lower quartile of the distribution, then that number is said to be an outlier.  

The lower quartile is [tex]Q_1[/tex] and the upper quartile of distribution is [tex]Q_3[/tex]

Then lower boundary is given by [tex]=Q_1-1.5\times IQR[/tex]

The Upper boundary is given by  [tex]=Q_3+1.5\times IQR[/tex]

If any number or value of the distribution is greater than [tex]Q_3+1.5\times IQR[/tex] then that is called Upper Outlier.

Similarly, if any value or number of the distribution is less than [tex]Q_1-1.5\times IQR[/tex] then that is called Lower Outlier.

So the answers in the boxes are given by:

Learn more about Outliers here -

https://brainly.com/question/3631910

#SPJ2

Answer:

Slope = 0

Step-by-step explanation:

Given the pair of points: (-3,2) (7,2)

Slope formula: y2-y1/x2-x1

2-2/7-(-3) = 0/10

0/10 = 0

Answer:

a. False

b. z = 8

Step-by-step explanation:

a. Equation answer:2 + 6 = 10 Real answer:2 + 6 = 8

b. 3 + 5 = 8

Answer:

T (3 , -6)   O ( -6 , -4) and M (-9 , 6)

Step-by-step explanation:

(x , y) -> (-x , -y)

Answer:

[tex]\cot(x)+\cot(\frac{\pi}{2}-x)[/tex]

[tex]\cot(x)+\tan(x)[/tex]

[tex]\frac{\cos(x)}{\sin(x)}+\frac{\sin(x)}{\cos(x)}[/tex]

[tex]\frac{1}{\sin(x)}(\cos(x)+\sin(x)\frac{\sin(x)}{\cos(x)})[/tex]

[tex]\csc(x)(\cos(x)+\sin(x)\frac{\sin(x)}{\cos(x)})[/tex]

[tex]\csc(x)[\frac{\cos(x)\cos(x)}{\cos(x)}+\sin(x)\frac{sin(x)}{\cos(x)}][/tex]

[tex]\csc(x)[\frac{\cos(x)\cos(x)+\sin(x)\sin(x)}{\cos(x)}][/tex]

[tex]\csc(x)[\frac{\cos^2(x)+\sin^2(x)}{\cos(x)}][/tex]

[tex]\csc(x)[\frac{1}{\cos(x)}][/tex]

[tex]\csc(x)[\sec(x)][/tex]

[tex]\csc(x)[\csc(\frac{\pi}{2}-x)][/tex]

[tex]\csc(x)\csc(\frac{\pi}{2}-x)[/tex]

Step-by-step explanation:

I'm going to use [tex]x[/tex] instead of [tex]\theta[/tex] because it is less characters for me to type.

I'm going to start with the left hand side and see if I can turn it into the right hand side.

[tex]\cot(x)+\cot(\frac{\pi}{2}-x)[/tex]

I'm going to use a cofunction identity for the 2nd term.

This is the identity: [tex]\tan(x)=\cot(\frac{\pi}{2}-x)[/tex] I'm going to use there.

[tex]\cot(x)+\tan(x)[/tex]

I'm going to rewrite this in terms of [tex]\sin(x)[/tex] and [tex]\cos(x)[/tex] because I prefer to work in those terms. My objective here is to some how write this sum as a product.

I'm going to first use these quotient identities: [tex]\frac{\cos(x)}{\sin(x)}=\cot(x)[/tex] and [tex]\frac{\sin(x)}{\cos(x)}=\tan(x)[/tex]

So we have:

[tex]\frac{\cos(x)}{\sin(x)}+\frac{\sin(x)}{\cos(x)}[/tex]

I'm going to factor out [tex]\frac{1}{\sin(x)}[/tex] because if I do that I will have the [tex]\csc(x)[/tex] factor I see on the right by the reciprocal identity:

[tex]\csc(x)=\frac{1}{\sin(x)}[/tex]

[tex]\frac{1}{\sin(x)}(\cos(x)+\sin(x)\frac{\sin(x)}{\cos(x)})[/tex]

[tex]\csc(x)(\cos(x)+\sin(x)\frac{\sin(x)}{\cos(x)})[/tex]

Now I need to somehow show right right factor of this is equal to the right factor of the right hand side.

That is, I need to show [tex]\cos(x)+\sin(x)\frac{\sin(x)}{\cos(x)}[/tex] is equal to [tex]\csc(\frac{\pi}{2}-x)[/tex].

So since I want one term I'm going to write as a single fraction first:

[tex]\cos(x)+\sin(x)\frac{\sin(x)}{\cos(x)}[/tex]

Find a common denominator which is [tex]\cos(x)[/tex]:

[tex]\frac{\cos(x)\cos(x)}{\cos(x)}+\sin(x)\frac{sin(x)}{\cos(x)}[/tex]

[tex]\frac{\cos(x)\cos(x)+\sin(x)\sin(x)}{\cos(x)}[/tex]

[tex]\frac{\cos^2(x)+\sin^2(x)}{\cos(x)}[/tex]

By  the Pythagorean Identity [tex]\cos^2(x)+\sin^2(x)=1[/tex] I can rewrite the top as 1:

[tex]\frac{1}{\cos(x)}[/tex]

By the quotient identity [tex]\sec(x)=\frac{1}{\cos(x)}[/tex], I can rewrite this as:

[tex]\sec(x)[/tex]

By the cofunction identity [tex]\sec(x)=\csc(x)=(\frac{\pi}{2}-x)[/tex], we have the second factor of the right hand side:

[tex]\csc(\frac{\pi}{2}-x)[/tex]

Let's just do it all together without all the words now:

[tex]\cot(x)+\cot(\frac{\pi}{2}-x)[/tex]

[tex]\cot(x)+\tan(x)[/tex]

[tex]\frac{\cos(x)}{\sin(x)}+\frac{\sin(x)}{\cos(x)}[/tex]

[tex]\frac{1}{\sin(x)}(\cos(x)+\sin(x)\frac{\sin(x)}{\cos(x)})[/tex]

[tex]\csc(x)(\cos(x)+\sin(x)\frac{\sin(x)}{\cos(x)})[/tex]

[tex]\csc(x)[\frac{\cos(x)\cos(x)}{\cos(x)}+\sin(x)\frac{sin(x)}{\cos(x)}][/tex]

[tex]\csc(x)[\frac{\cos(x)\cos(x)+\sin(x)\sin(x)}{\cos(x)}][/tex]

[tex]\csc(x)[\frac{\cos^2(x)+\sin^2(x)}{\cos(x)}][/tex]

[tex]\csc(x)[\frac{1}{\cos(x)}][/tex]

[tex]\csc(x)[\sec(x)][/tex]

[tex]\csc(x)[\csc(\frac{\pi}{2}-x)][/tex]

[tex]\csc(x)\csc(\frac{\pi}{2}-x)[/tex]

Answer:

multiply both numbers and if im correct it should be 124ft sorry if im wrong

Answer:

y = 5

Step-by-step explanation:

The y-axis and any lines that are parallel to it are vertical. Therefore, a line that would be perpendicular to the y-axis would need to be horizontal. All horizontal lines have a slope equal to 0 and are represented by the equation y = a constant, or a single number with no variables.  

The only option that follows that format is y = 5, thus that is the answer.  

Casually puts a link to a weird website..

https://theuselessweb.com

*whistles*

Answer:

there is no solution-

Step-by-step explanation:

there's no solution because its impossible for a to have 2 different values; you see that 0.1 and 0.01 are not equal, therefore a can not be equal,, otherwise the equation would be false

Answer:

i can.

Step-by-step explanation: