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Umnica [9.8K]
3 years ago
14

Iven triangle abc and def which of the following statements must be true LMK ASAP

Mathematics
1 answer:
VLD [36.1K]3 years ago
4 0
A. would be true, but double check someone?
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A robin can fly 10 miles per hour faster than a blue jay. It takes a robin 1 1/2 hours to fly 45 miles.
cricket20 [7]

Answer: 2\dfrac14\text{ hours}

Step-by-step explanation:

Given:  It takes a robin 1\dfrac12 hours to fly 45 miles.

1\dfrac12=\dfrac32

Speed of robin= \dfrac{distance}{time}=\dfrac{45}{\dfrac32}

\dfrac{45}{3}\times2= 15\times2= 30\text{ miles per hour}

A robin can fly 10 miles per hour faster than a blue jay. It

Speed of Blue jay = Speed of robin - 10 miles per hour

= 30 miles per hour - 10 miles per hour

= 20 miles per hour

Time taken to cover 45 miles = \dfrac{45}{20}=\dfrac{9}{4}=2\dfrac14\text{ hours}

Hence, it will take 2\dfrac14\text{ hours}  by a blue jay to fly that same distance.

4 0
3 years ago
What is the solution to the following system of equations?
lisov135 [29]

Answer:

d

4*2 + 2*-1= 6

2- -1= 3

4 0
3 years ago
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Verify that:
Lelu [443]

Answer:

See Below.

Step-by-step explanation:

Problem 1)

We want to verify that:

\displaystyle \left(\cos(x)\right)\left(\cot(x)\right)=\csc(x)-\sin(x)

Note that cot(x) = cos(x) / sin(x). Hence:

\displaystyle \left(\cos(x)\right)\left(\frac{\cos(x)}{\sin(x)}\right)=\csc(x)-\sin(x)

Multiply:

\displaystyle \frac{\cos^2(x)}{\sin(x)}=\csc(x)-\sin(x)

Recall that Pythagorean Identity: sin²(x) + cos²(x) = 1 or cos²(x) = 1 - sin²(x). Substitute:

\displaystyle \frac{1-\sin^2(x)}{\sin(x)}=\csc(x)-\sin(x)

Split:

\displaystyle \frac{1}{\sin(x)}-\frac{\sin^2(x)}{\sin(x)}=\csc(x)-\sin(x)

Simplify:

\csc(x)-\sin(x)=\csc(x)-\sin(x)

Problem 2)

We want to verify that:

\displaystyle (\csc(x)-\cot(x))^2=\frac{1-\cos(x)}{1+\cos(x)}

Square:

\displaystyle \csc^2(x)-2\csc(x)\cot(x)+\cot^2(x)=\frac{1-\cos(x)}{1+\cos(x)}

Convert csc(x) to 1 / sin(x) and cot(x) to cos(x) / sin(x). Thus:

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

Factor out the sin²(x) from the denominator:

\displaystyle \frac{1}{\sin^2(x)}\left(1-2\cos(x)+\cos^2(x)\right)=\frac{1-\cos(x)}{1+\cos(x)}

Factor (perfect square trinomial):

\displaystyle \frac{1}{\sin^2(x)}\left((\cos(x)-1)^2\right)=\frac{1-\cos(x)}{1+\cos(x)}

Using the Pythagorean Identity, we know that sin²(x) = 1 - cos²(x). Hence:

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

Factor (difference of two squares):

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

Factor out a negative from the first factor in the denominator:

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

Cancel:

\displaystyle \frac{\cos(x)-1}{-(1+\cos(x))}=\frac{1-\cos(x)}{1+\cos(x)}

Distribute the negative into the numerator. Therefore:

\displaystyle \frac{1-\cos(x)}{1+\cos(x)}=\displaystyle \frac{1-\cos(x)}{1+\cos(x)}

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3 years ago
I REALLY NEED HELP ON THIS. It's due tomorrow. please help
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Question fourteen is ten but thats all i got
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3 years ago
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Please help can give brainliest
DIA [1.3K]

Answer:

you have to add them together

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