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3 years ago

Does anybody know this

Mathematics

1 answer:

Makovka662 [10]3 years ago

7 0

Answer:

1) 75°

3) 135°

4) 105°

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What is the geometric mean of 3 and 15?

7nadin3 [17]

√(3*15)=√45=√[5*3^(2)]=3√5

3 0

3 years ago

the length of a rectangle is five times its width. if the area of the rectangle is 245 meters squared, find its perimeter

Vikentia [17]

Your answer is 84 meters.

We can call the width x, and thus the length 5x, and then we can say that 5x × x = 245, which simplifies to 5x² = 245, and then we can solve:

5x² = 245

÷ 5

x² = 49

√

x = 7

Now that we know what x is, we can find the perimeter by using it as the width, and then multiplying it by 5 to find its length. This gives us:

(7 × 2) + (5 × 7 × 2) = 14 + (35 × 2) = 14 + 70 = 84

So your answer is 84 meters.

I hope this helps!

6 0

3 years ago

6 x to the 3 fourths power in radical form? (1 point) Question 3 options: 1) cubed root of 6 x to the fourth power 2) fourth roo

vivado [14]

Answer:4) 6 fourth root of x cubed

Step-by-step explanation:

6 0

3 years ago

Pleaseee help me....what will the y-coordinate of b' if ab is dilated by a factor of 2/5, centered at the origin. A=(2,2) B=(10,

sdas [7]

Answer:

The y-coordinate of B' is equal to 2

Step-by-step explanation:

step 1

Find the distance of AB

$AB=10-2=8\ units$

step 2

Find the length of A'B'

Multiply the length of AB by the scale factor

$A'B'=8*\frac{2}{5}= 3.2\ units$

step 3

Find the midpoint AB

$M=(\frac{2+10}{2},\frac{2+2}{2} )$

$M=(6,2)$

step 4

<em>Find the x-coordinate of B'</em>

The x-coordinate of B' is equal to the x-coordinate of midpoint plus the semi distance A'B'

$6+\frac{3.2}{2}=6+1.6=7.6$

<em>Find the y-coordinate of B'</em>

The y-coordinate of B' is equal to 2 (because is a horizontal line)

3 0

3 years ago

Read 2 more answers

If tanA+cotA=2 then find the value of cosA

aniked [119]

Answer:

cos A=

$\frac{1}{ \sqrt{2} }$

Step-by-step explanation:

We know,

cotA=

$\frac{1}{ \tan(a) }$

Substituting the value of cot A in the given equation, we get

$\tan( a) + \frac{1}{ \tan(a) } = 2$

$\frac{{ \tan }^{2} a + 1}{ \: \tan(a) } = 2$

${ \tan }^{2 \: } a \: + 1 = 2 tan \: a$

${ \tan }^{2} a - 2 \tan \: a + 1 = 0$

$( \tan \: a - 1) {}^{2 } = 0$

$\tan \: a - 1 = \sqrt{0}$

$\tan \: a = 1$

$\tan \: a \: = \tan \: 45$

$a = 45$

$\cos \: a = \cos \: 45$

$\cos \: a = \frac{1}{ \sqrt{2} }$

4 0

2 years ago