All categories

3 years ago

ASAP PLEASE HELP!! SCALE FACTOR! EASY!!

Mathematics

2 answers:

zysi [14]3 years ago

6 0

Answer:

1.6667

Step-by-step explanation:

professor190 [17]3 years ago

6 0

I think it’s 1.666 I’m not sure

You might be interested in

Hi! Can anyone with knowledge on writing Geometry proofs help me? I need assistance with answering this question with a two colu

Neporo4naja [7]

We will prove first that PRQ≡PSQ
We need 3 attributes from the each triangle:
QS=QR
PR=PS
PQ=PQ

from PRQ≡PSQ => angles RPQ and SPQ are equal

Let if be O=RS∩PQ

We will prove that triangles RPO≡SPO
We need 3 attributes from the each triangle:
RP=PS
PO=PO
angles RPQ and SPQ are equal

results that angles RPO and SPO are equal(Let's note them with x)
We also know that (as angles) POS≡ROS and POR≡SOQ as opposite angles

We now have than 4x=360 wich means x=90. And we proved the desired result.
You just need to put this explanation into the table. (most if the 3 attributes for each triangle)

3 0

3 years ago

From the set {1, 3, 4}, use substitution to determine which value of x makes the equation true.

Harlamova29_29 [7]

30 - 2x= 22

-2x = 22 - 30

-2x = -8

(dived both sides by -2)

x= 4

8 0

3 years ago

You are watching a commercial on television for the newest sandwich at ur favorite fast food chain.what type of promotion are u

zheka24 [161]

You are watching an advertisement

4 0

3 years ago

Read 2 more answers

Help please ITS OF TRIGONOMETRY PROVE

Flauer [41]

Answer:

The equation is true.

Step-by-step explanation:

In order to solve this problem, one must envision a right triangle. A diagram used to represent the imagined right triangle is included at the bottom of this explanation. Please note that each side is named with respect to the angle is it across from.

Right angle trigonometry is composed of a sequence of ratios that relate the sides and angles of a right triangle. These ratios are as follows,

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

One is given the following equation,

$\frac{sin(A)+sin(B)}{cos(A) +cos(B)}+\frac{cos(A)-cos(B)}{sin(A)-sin(B)}=0$

As per the attached reference image, one can state the following, using the right angle trigonometric ratios,

$sin(A)=\frac{a}{c}\\\\sin(B)=\frac{b}{c}\\\\cos(A)=\frac{b}{c}\\\\cos(B)=\frac{a}{c}$

Substitute these values into the given equation. Then simplify the equation to prove the idenity,

$\frac{sin(A)+sin(B)}{cos(A) +cos(B)}+\frac{cos(A)-cos(B)}{sin(A)-sin(B)}=0$

$\frac{\frac{a}{c}+\frac{b}{c}}{\frac{b}{c}+\frac{a}{c}}+\frac{\frac{b}{c}-\frac{a}{c}}{\frac{a}{c}-\frac{b}{c}}=0$

$\frac{\frac{a+b}{c}}{\frac{a+b}{c}}+\frac{\frac{b-a}{c}}{\frac{a-b}{c}}$

Remember, any number over itself equals one, this holds true even for fractions with fractions in the numerator (value on top of the fraction bar) and denominator (value under the fraction bar).

$\frac{\frac{a+b}{c}}{\frac{a+b}{c}}+\frac{\frac{b-a}{c}}{\frac{a-b}{c}}$

$\frac{\frac{a+b}{c}}{\frac{a+b}{c}}+\frac{\frac{-(a-b)}{c}}{\frac{a-b}{c}}$