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Allushta [10]
3 years ago
13

Units needed 90 units per case 8

Mathematics
2 answers:
True [87]3 years ago
6 0

Answer:

units needed 90 units per case 8 is

90/8=11.25

S_A_V [24]3 years ago
5 0
What is the question?
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Hi there!

There are many ways to find the measure of an exterior angle of a triangle. I'll explain the easiest method to you. As you can see in the image, a line is drawn extending a side of the triangle. In order to find the measure of the exterior angle, we can subtract the measure of the adjacent interior angle from 180 (a straight angle). Doing this subtraction will give us the measure of the exterior angle.

Hope this helps!! :)

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4 years ago
A line is defined by the equation y = negative x + 3. Which shows the graph of this line? On a coordinate plane, a line goes thr
andrew-mc [135]

Answer:

A

A line is defined by the equation y =

negative x + 3

. Which shows the graph of this line? On a coordinate plane, a line goes

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On a coordinate plane, a line goes through points (0, negative 3) and (0, 3).

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goes through points (negative 2, 6) and (0,

5 0
3 years ago
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2(3p+4)–<br> 2<br> 3<br> p=<br> 1<br> 3<br> (9+p)<br> p=
kobusy [5.1K]

Answer:

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3 0
3 years ago
Write a polynomial f(x) that satisfies the given conditions.
patriot [66]

Answer:

f(x) = (x-4)(x-\frac{1}{2})^2 = x^3-5x^2+4.25x -1

Step-by-step explanation:

Recall, if we have a polynomial of the form (x-a)^k \cdot (x-b)^m, then we say that a is a zero of multiplicity k and b is a zero of multiplicty m. For example, in the polynomial of the form (x+5)^10(x-2)^3 -5 is a zero of multiplicity 10 and 2 is a zero of multiplicity 3. If we want to know the degree of the polynomial,  just add the multiplicity of both zeros (13 in our example).

In this case, we know that the degree of our polynomial should be at least 3(multiplicity 2 and multiplicity 1). So, lets take the polynomial of the form

f(x)=(x-a)(x-b)^2).

In here, a is a zero with multiplicity 1 and b is a zero with multiplicity 2. We are also given that

f(0) = -1 = (0-a)(0-b)^2 = -a\codt b^2.

Which implies that 1 = a\cdot b^2. Since the square of any number is a positive number, it must happen that a>0. So, we have that

b= \pm \sqrt[]{\frac{1}{a}}.

We can choose any value of a and solve for b. Let us choose a=4. So we can have b=1/2 or b=-1/2. Let's use b=1/2. So our polynomial would be

f(x) = (x-4)(x-\frac{1}{2})^2 = x^3-5x^2+4.25x -1

which we can easily check that f(0)=-1.

6 0
4 years ago
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