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tino4ka555 [31]
3 years ago
11

Please help with math question

Mathematics
1 answer:
Pachacha [2.7K]3 years ago
5 0
I can try and make it easier for you. The first equation is already in a form that can be easily graphed.

y = -5x + 3

I will make the second one into the same form.

y = -5x + 3 (This is also the second line)

Essentially, a solution for this system of equations is going to be any point on the line, because they are identical.

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Konrad raised $75.50 for his football team. He raised 151% of his fundraising goal how much more then his fundraising did Konrad
Colt1911 [192]

Answer:

He raised $25.50 over his fund raising goal

Step-by-step explanation:

Here, we want to calculate his fund raising goal.

Let his fund raising goal be $x

We are told that he raised 151% of his fundraising goal;

Thus,

mathematically;

151% of x = 75.5

151/100 * x = 75.5

151x = 100 * 75.5

151x = 7550

x = 7550/151

x = $50

So the amount he raised over his fund raising goal will be $75.50 - $50 = $25.50

8 0
3 years ago
8. 36.8 = [11.5 – (2.5 X 3)]2
Anestetic [448]

Answer:

[11.5-(2.5*3)]^2=16

Step-by-step explanation:

We want to evaluate [11.5-(2.5*3)]^2

Let us evaluate within the parenthesis first:

[11.5-(2.5*3)]^2=[11.5-(7.5)]^2

\implies [11.5-(2.5*3)]^2=[11.5-7.5]^2

We again subtract within the bracket to obtain:

[11.5-(2.5*3)]^2=[4]^2

This finally gives us:

[11.5-(2.5*3)]^2=16

6 0
2 years ago
Why is it possible to have two different rational functions with the same asymptotes?
jolli1 [7]

Answer: Down below.

Step-by-step explanation:

A horizontal asymptote for a function is a horizontal line that the graph of the function approaches as x approaches ∞ (infinity) or -∞ (minus infinity). ... There are literally only two limits to look at, so that means there can only be at most two horizontal asymptotes for a given function

4 0
3 years ago
Given that cos θ = –12∕13 and θ is in quadrant III, find csc θ.
Trava [24]

Answer:

A

Step-by-step explanation:

We are given:

\displaystyle \cos(\theta)=-\frac{12}{13},\,\theta\in\text{QIII}

Since cosine is the ratio of the adjacent side over the hypotenuse, this means that the opposite side is (we can ignore negatives for now):

o=\sqrt{13^2-12^2}=\sqrt{25}=5

So, the opposite side is 5, the adjacent side is 12, and the hypotenuse is 13.

And since θ is in QIII, sine/cosecant is negative, cosine/secant is negative, and tangent/cotangent is positive.

Cosecant is given by the hypotenuse over the opposite side. Thus:

\displaystyle \csc(\theta)=\frac{13}{5}

Since θ is in QIII, cosecant must be negative:

\displaystyle \csc(\theta)=-\frac{13}{5}

Our answer is A.

7 0
2 years ago
Read 2 more answers
(5⁴)²?????? <br><br><br>A:5⁶<br>B:5⁸<br>C:20²<br>D:25⁸​
ArbitrLikvidat [17]

Answer:c 20

Step-by-step explanation:

4 0
3 years ago
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