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2 years ago

Someone please help!!

Mathematics

1 answer:

goldenfox [79]2 years ago

3 0

Reflection, the quadratic parent function is reflected on the opposite side

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Mrs. Wright bought some notebook paper for her class. She decided to keep 3/4 of the paper for her own use. How much paper did s

larisa [96]

So the total amount of paper is 4/4 right?
And she kept 3/4.
So we have to minus 3/4 from 4/4.
Because it has the same denominator we need not change it to similar one.So 4-3 is one.
So answer is 1/4

5 0

3 years ago

Given that these simultaneous equations

Tatiana [17]

The value of k is $(3\sqrt{ 10})/2$

<h3>How to solve the simultaneous equation?</h3>

Given:

x-y=k.............(eq i)

2x²+y²-15..............(eq ii)

We would make y the subject formula in eq ii

2x²+y²-15= 0

2x² + y²= 15

y²= 15-2x²

y= $\sqrt{15-2x^2}$ ...........(eq iii)

Substitute the value of y into eq i

x-( $\sqrt{15-2x^2}$ = k

x- ( $\sqrt{15} - 2x$ = k

k= $(3\sqrt{ 10})/2$

Read more about simultaneous equations here:

brainly.com/question/16863577

#SPJ1

3 0

1 year ago

What is the period of y = 8 cos (5pie x + 3pie/2)-9<br> Give an exact value. Units

zzz [600]

$\\ \rm\Rrightarrow y=Acos(Bx+C)+D$

This is the general formula and the period is 2π/B

Now our given function

$\\ \rm\Rrightarrow y=8cos\left(5\pi x+\dfrac{3\pi}{2}\right)-9$

On comparing we get

A=8
B=5π
C=3π/2
D=-9

Period:-

$\\ \rm\Rrightarrow \dfrac{2\pi}{B}$

$\\ \rm\Rrightarrow \dfrac{2\pi}{5\pi}$

$\\ \rm\Rrightarrow \dfrac{2}{5}$

3 0

2 years ago

Read 2 more answers

Which is a correct way to find f(3) when f(x) = 2x27

dalvyx [7]

The answer to = 2x27 is f(x)

5 0

2 years ago

Please give me correct answer and fast answer it if know answer only

Stolb23 [73]

Answer:

Approximatley 5.8 units.

Step-by-step explanation:

We are given two angles, ∠S and ∠T, and the side opposite to ∠T. We need to find the unknown side opposite to ∠S. Therefore, we can use the Law of Sines. The Law of Sines states that:

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

Replacing them with the respective variables, we have:

$\frac{\sin(S)}{s} =\frac{\sin(T)}{t} =\frac{\sin(R)}{r}$

Plug in what we know. 20° for ∠S, 17° for ∠T, and 5 for <em>t</em>. Ignore the third term:

$\frac{\sin(20)}{s}=\frac{\\sin(17)}{5}$

Solve for <em>s</em>, the unknown side. Cross multiply:

$\frac{\sin(20)}{s}=\frac{\sin(17)}{5}\\5\sin(20)=s\sin(17)\\s=\frac{5\sin(20)}{\sin(17)} \\s\approx5.8491\approx5.8$

6 0

3 years ago

Read 2 more answers