All categories

3 years ago

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?

You might be interested in

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:

4*2 + 2*-1= 6

2- -1= 3

4 0

3 years ago

Read 2 more answers

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:

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

3 0

3 years ago

I REALLY NEED HELP ON THIS. It's due tomorrow. please help

lys-0071 [83]

Question fourteen is ten but thats all i got

6 0

3 years ago

Read 2 more answers

Please help can give brainliest

DIA [1.3K]

Answer:

you have to add them together

4 0

3 years ago

Read 2 more answers