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3 years ago

30 pts!!!

Mathematics

1 answer:

Nady [450]3 years ago

4 0

A and b are equivalent to 34

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50 points. <br> Solve the following problems.

jeyben [28]

Answer:

x=10, y=12.

Step-by-step explanation:

You are making all sides of the triangle smaller by a scale factor of 2/3, because 9*2/3=6. For x, we can do 2/3*15, which equals 10. For y, we can do 2/3*18, equalling 12.

3 0

3 years ago

Read 2 more answers

If you want Brainliest! You gotta earn it by obviously helping! I need help ASAP, please!

masha68 [24]

Answer:

$\theta=1.72$ radians

Step-by-step explanation:

Arc length and radius is given. So, we need to know the arc length formula for a circle.

That is:

$s=r\theta$

Where s is the arc length

r is the radius

$\theta$ is the angle (IN RADIANS)

We have:

s = 15.5

r = 9

Now, we substitute and find the angle:

$s=r\theta\\15.5=9\theta\\\theta=\frac{15.5}{9}\\\theta=1.72$

The angle, in radians, in 1.72

5 0

3 years ago

A rectangle is transformed according to the rule r0, 90º. the image of the rectangle has vertices located at r'(–4, 4), s'(–4, 1

Korvikt [17]

Answer:

Clockwise vertices of q' ( -3 ,4) →→ q( 4 , 3 ).

Counter clock wise rule : q' (-3 ,4 ) →→ q( -4 , -3 ).

Step-by-step explanation:

Given : rectangle has vertices located at r'(–4, 4), s'(–4, 1), p'(–3, 1), and q'(–3, 4)

To find : transformed according to the rule 90º , what is the location of q?

Solution : we have given that

vertices located at r'(–4, 4), s'(–4, 1), p'(–3, 1), and q'(–3, 4).

By the rule of 90º rotation clock wise rule : (x ,y ) →→ ( y , -x )

90º rotation counter clock wise rule : (x ,y ) →→ ( -y , x ).

Then Clockwise vertices of q' ( -3 ,4) →→ q( 4 , 3 ).

counter clock wise rule : q' (-3 ,4 ) →→ q( -4 , -3 ).

Therefore, Clockwise vertices of q' ( -3 ,4) →→ q( 4 , 3 ).

counter clock wise rule : q' (-3 ,4 ) →→ q( -4 , -3 ).

4 0

3 years ago

Read 2 more answers

What is the mean? <br> 19, 16, 24, 12, 16, 21, 18

attashe74 [19]

I'm pretty sure it's 12.

6 0

3 years ago

Read 2 more answers

PLS ANSWER ASAP 30 POINTS!!! CHECK PHOTO! WILL MARK BRAINLIEST TO WHO ANSWERS

Sveta_85 [38]

I'll do Problem 8 to get you started

a = 4 and c = 7 are the two given sides

Use these values in the pythagorean theorem to find side b

$a^2 + b^2 = c^2\\\\4^2 + b^2 = 7^2\\\\16 + b^2 = 49\\\\b^2 = 49 - 16\\\\b^2 = 33\\\\b = \sqrt{33}\\\\$

With respect to reference angle A, we have:

opposite side = a = 4
adjacent side = b = $\sqrt{33}$
hypotenuse = c = 7

Now let's compute the 6 trig ratios for the angle A.

We'll start with the sine ratio which is opposite over hypotenuse.

$\sin(\text{angle}) = \frac{\text{opposite}}{\text{hypotenuse}}\\\\\sin(A) = \frac{a}{c}\\\\\sin(A) = \frac{4}{7}\\\\$

Then cosine which is adjacent over hypotenuse

$\cos(\text{angle}) = \frac{\text{adjacent}}{\text{hypotenuse}}\\\\\cos(A) = \frac{b}{c}\\\\\cos(A) = \frac{\sqrt{33}}{7}\\\\$

Tangent is the ratio of opposite over adjacent

$\tan(\text{angle}) = \frac{\text{opposite}}{\text{adjacent}}\\\\\tan(A) = \frac{a}{b}\\\\\tan(A) = \frac{4}{\sqrt{33}}\\\\\tan(A) = \frac{4\sqrt{33}}{\sqrt{33}*\sqrt{33}}\\\\\tan(A) = \frac{4\sqrt{33}}{(\sqrt{33})^2}\\\\\tan(A) = \frac{4\sqrt{33}}{33}\\\\$

Rationalizing the denominator may be optional, so I would ask your teacher for clarification.

So far we've taken care of 3 trig functions. The remaining 3 are reciprocals of the ones mentioned so far.

cosecant, abbreviated as csc, is the reciprocal of sine
secant, abbreviated as sec, is the reciprocal of cosine
cotangent, abbreviated as cot, is the reciprocal of tangent

So we'll flip the fraction of each like so:

$\csc(\text{angle}) = \frac{\text{hypotenuse}}{\text{opposite}} \ \text{ ... reciprocal of sine}\\\\\csc(A) = \frac{c}{a}\\\\\csc(A) = \frac{7}{4}\\\\\sec(\text{angle}) = \frac{\text{hypotenuse}}{\text{adjacent}} \ \text{ ... reciprocal of cosine}\\\\\sec(A) = \frac{c}{b}\\\\\sec(A) = \frac{7}{\sqrt{33}} = \frac{7\sqrt{33}}{33}\\\\\cot(\text{angle}) = \frac{\text{adjacent}}{\text{opposite}} \ \text{ ... reciprocal of tangent}\\\\\cot(A) = \frac{b}{a}\\\\\cot(A) = \frac{\sqrt{33}}{4}\\\\$

------------------------------------------------------

Summary:

The missing side is $b = \sqrt{33}$

The 6 trig functions have these results

$\sin(A) = \frac{4}{7}\\\\\cos(A) = \frac{\sqrt{33}}{7}\\\\\tan(A) = \frac{4}{\sqrt{33}} = \frac{4\sqrt{33}}{33}\\\\\csc(A) = \frac{7}{4}\\\\\sec(A) = \frac{7}{\sqrt{33}} = \frac{7\sqrt{33}}{33}\\\\\cot(A) = \frac{\sqrt{33}}{4}\\\\$

Rationalizing the denominator may be optional, but I would ask your teacher to be sure.

7 0

2 years ago