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3 years ago

Plz, HELP me with this question!

Mathematics

1 answer:

Tresset [83]3 years ago

7 0

1/6 should be the right answer

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How is the scale factor, surface area ratio and volume ratio related?

Travka [436]

The ratio of the surface areas of similar solids is equal to the square of their scale factor and that the ratio of their volumes is equal to the cube of their scale factor.

5 0

1 year ago

Hi can anybody explain to me how to do this I'm kinda confused . 5×[5(-7)]

telo118 [61]

Answer:

5 × [5(-7)]

= 5 × (-35) ---- 5 × (-7) = -35

= -175

4 0

3 years ago

PLEASE HELP :)))<br><br> triangle similarity theorem

Harman [31]

Answer:

Step-by-step explanation

3 0

3 years ago

Find exact values for sin θ, cos θ and tan θ if csc θ = 3/2 and cos θ < 0.

kumpel [21]

Answer:

Part 1) $sin(\theta)=\frac{2}{3}$

Par 2) $cos(\theta)=-\frac{\sqrt{5}}{3}$

Part 3) $tan(\theta)=-\frac{2\sqrt{5}}{5}$

Step-by-step explanation:

step 1

Find the $sin(\theta)$

we have

$csc(\theta)=\frac{3}{2}$

Remember that

$csc(\theta)=\frac{1}{sin(\theta)}$

therefore

$sin(\theta)=\frac{2}{3}$

step 2

Find the $cos(\theta)$

we know that

$sin^{2}(\theta) +cos^{2}(\theta)=1$

we have

$sin(\theta)=\frac{2}{3}$

substitute

$(\frac{2}{3})^{2} +cos^{2}(\theta)=1$

$\frac{4}{9} +cos^{2}(\theta)=1$

$cos^{2}(\theta)=1-\frac{4}{9}$

$cos^{2}(\theta)=\frac{5}{9}$

square root both sides

$cos(\theta)=\pm\frac{\sqrt{5}}{3}$

we have that

$cos(\theta) < 0$ ---> given problem

$cos(\theta)=-\frac{\sqrt{5}}{3}$

step 3

Find the $tan(\theta)$

we know that

$tan(\theta)=\frac{sin(\theta)}{cos(\theta)}$

we have

$sin(\theta)=\frac{2}{3}$

$cos(\theta)=-\frac{\sqrt{5}}{3}$

substitute

$tan(\theta)=\frac{2}{3}:-\frac{\sqrt{5}}{3}=-\frac{2}{\sqrt{5}}$

Simplify

$tan(\theta)=-\frac{2\sqrt{5}}{5}$

6 0

2 years ago

Show that if u+v and u-v are orthognal, then the vectors u and v must have the same length.

pashok25 [27]

Answer with Step-by-step explanation:

We are given that

u+ v and u-v are orthogonal

We have to prove that u and v must have the same length.

When two vector a and b are orthogonal then

$a\cdot b=0$

By using the property

$(u+v)\cdot (u-v)=0$

We know that

$(a+b)\cdot (a-b)=\mid a\mid^2-\mid b\mid^2$

$\mid u\mid ^2-\mid v\mid^2=0$

$\mid u\mid^2=\mid v\mid^2$

Magnitude is always positive

When power of base on both sides are equal then base will be equal

Therefore,

$\mid u\mid=\mid v\mid$

Hence, the length of vectors u and v must have the same length.

5 0

3 years ago