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2 years ago

Plsss helppp! i will give brainliest!!

Mathematics

2 answers:

algol132 years ago

8 0

U need to find the graph

7nadin3 [17]2 years ago

6 0

It’s option a. The question is a bit tricky because you’re given the image and asked to find the pre image.

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What is the slope of the line? pls help

Flauer [41]

Answer:Rise= 4 Run= 5

Step-by-step explanation:

You have to do Rise over Run. You have to find two of the points and count going up and right.

3 0

2 years ago

What is the value of 3 + (-1/4)² ÷ 1/2? A. 6 1/8 B. 21 1/8 C. 7 D. 3 1/8

Amiraneli [1.4K]

3 + (-1/4)² ÷ 1/2
= 3 + 1/16 * 2
= 3 + 1/8
= 3 1/8

answer

D. 3 1/8

6 0

3 years ago

An investor invested a total of $2,800 in two mutual funds. One fund earned 5% profit while the other earned 3% profit. If the i

Natasha_Volkova [10]

Answer:

The amount invested in the mutual fund that earned 5% was $1,000

The amount invested in the mutual fund that earned 3% was $1,800

Step-by-step explanation:

Let

x ----> the amount invested in the mutual fund that earned 5%

y ----> the amount invested in the mutual fund that earned 3%

we know that

$x+y=2,800$ ----> $y=2,800-x$ ----> equation A

$5\%=5/100=0.05$

$3\%=3/100=0.03$

$0.05x+0.03y=104$ ----> equation B

Solve the system by substitution

substitute equation A in equation B

$0.05x+0.03(2,800-x)=104$

Solve for x

$0.05x+84-0.03x=104$

$0.05x-0.03x=104-84$

$0.02x=20$

$x=\$1,000$

Find the value of y

$y=2,800-x$

$y=2,800-1,000=\$1,800$

therefore

The amount invested in the mutual fund that earned 5% was $1,000 and the amount invested in the mutual fund that earned 3% was $1,800

5 0

3 years ago

I need help if anyone can help me with a few problems please help me

jok3333 [9.3K]

I think it’s the number 3 I’m not sure I’m learning this stuff rn I’m getting the hang of it just it’s hard and not hard at the same time

5 0

3 years ago

Show tan(???? − ????) = tan(????)−tan(????) / 1+tan(????) tan(????) .

anyanavicka [17]

Answer:

See the proof below

Step-by-step explanation:

For this case we need to proof the following identity:

$tan(x-y) = \frac{tan(x) -tan(y)}{1+ tan(x) tan(y)}$

We need to begin with the definition of tangent:

$tan (x) =\frac{sin(x)}{cos(x)}$

So we can replace into our formula and we got:

$tan(x-y) = \frac{sin(x-y)}{cos(x-y)}$ (1)

We have the following identities useful for this case:

$sin(a-b) = sin(a) cos(b) - sin(b) cos(a)$

$cos(a-b) = cos(a) cos(b) + sin (a) sin(b)$

If we apply the identities into our equation (1) we got:

$tan(x-y) = \frac{sin(x) cos(y) - sin(y) cos(x)}{sin(x) sin(y) + cos(x) cos(y)}$ (2)

Now we can divide the numerator and denominato from expression (2) by $\frac{1}{cos(x) cos(y)}$ and we got this:

$tan(x-y) = \frac{\frac{sin(x) cos(y)}{cos(x) cos(y)} - \frac{sin(y) cos(x)}{cos(x) cos(y)}}{\frac{sin(x) sin(y)}{cos(x) cos(y)} +\frac{cos(x) cos(y)}{cos(x) cos(y)}}$

And simplifying we got:

$tan(x-y) = \frac{tan(x) -tan(y)}{1+ tan(x) tan(y)}$

And this identity is satisfied for all:

$(x-y) \neq \frac{\pi}{2} +n\pi$

8 0

3 years ago