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galben [10]
3 years ago
6

Can someone help me on this problem

Mathematics
2 answers:
Svetradugi [14.3K]3 years ago
5 0
1.5p+3p+2.5p= total amount
alukav5142 [94]3 years ago
4 0
You will earn 7 dollars each person
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✓
ikadub [295]

Answer:

the corresponding sides and angles would be a pair of matching angles or sides that are in the same spot in two different shapes.

Step-by-step explanation:

So look at triangle AWXV and check each side....does side AW correspond with other triangle AX - meaning they are the same?

Repeat to find the same angle in each triangle that matches.

8 0
3 years ago
Which of the following most commonly influences weather?
astraxan [27]
The answer should be jet streams. :)
6 0
3 years ago
) f) 1 + cot²a = cosec²a​
notsponge [240]

Answer:

It is an identity, proved below.

Step-by-step explanation:

I assume you want to prove the identity. There are several ways to prove the identity but here I will prove using one of method.

First, we have to know what cot and cosec are. They both are the reciprocal of sin (cosec) and tan (cot).

\displaystyle \large{\cot x=\frac{1}{\tan x}}\\\displaystyle \large{\csc x=\frac{1}{\sin x}}

csc is mostly written which is cosec, first we have to write in 1/tan and 1/sin form.

\displaystyle \large{1+(\frac{1}{\tan x})^2=(\frac{1}{\sin x})^2}\\\displaystyle \large{1+\frac{1}{\tan^2x}=\frac{1}{\sin^2x}}

Another identity is:

\displaystyle \large{\tan x=\frac{\sin x}{\cos x}}

Therefore:

\displaystyle \large{1+\frac{1}{(\frac{\sin x}{\cos x})^2}=\frac{1}{\sin^2x}}\\\displaystyle \large{1+\frac{1}{\frac{\sin^2x}{\cos^2x}}=\frac{1}{\sin^2x}}\\\displaystyle \large{1+\frac{\cos^2x}{\sin^2x}=\frac{1}{\sin^2x}}

Now this is easier to prove because of same denominator, next step is to multiply 1 by sin^2x with denominator and numerator.

\displaystyle \large{\frac{\sin^2x}{\sin^2x}+\frac{\cos^2x}{\sin^2x}=\frac{1}{\sin^2x}}\\\displaystyle \large{\frac{\sin^2x+\cos^2x}{\sin^2x}=\frac{1}{\sin^2x}

Another identity:

\displaystyle \large{\sin^2x+\cos^2x=1}

Therefore:

\displaystyle \large{\frac{\sin^2x+\cos^2x}{\sin^2x}=\frac{1}{\sin^2x}\longrightarrow \boxed{ \frac{1}{\sin^2x}={\frac{1}{\sin^2x}}}

Hence proved, this is proof by using identity helping to find the specific identity.

6 0
3 years ago
Plz help ASAP!!!!!!!!!<br><br> thanks
iogann1982 [59]

Answer:

D (Last one)

Step-by-step explanation:

Reflections and rotations dont change the side lengths or angles of a shape, dilations do.

7 0
3 years ago
What is the true solution to the equation below? 2 in e in2×-in e in 10×= in 30 A x=30 B x=75 C x=150 D x=300
hodyreva [135]

Answer:

Option B.

Step-by-step explanation:

Let as consider the given equation:

2\ln e^{\ln 2x}-\ln e^{\ln 10x}=\ln 30

It can be written as

2(\ln 2x)-(\ln 10x)=\ln 30         [\because \ln e^a=a]

\ln (2x)^2-(\ln 10x)=\ln 30        [\because \ln a^b=b\ln a]

\ln \dfrac{4x^2}{10x}=\ln 30        [\because \ln \dfrac{a}{b}=\ln a-\ln b]

\ln \dfrac{2x}{5}=\ln 30

On comparing both sides, we get

\dfrac{2x}{5}=30

Multiply both sides by 5.

2x=150

Divide both sides by 2.

x=75

Therefore, the correct option is B.

7 0
3 years ago
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