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Lubov Fominskaja [6]
3 years ago
10

Marvin scored 96, 93, 97, and 92 on four out of five math tests. In order to receive the highest grade in the class, he must obt

ain an average score of at least 95 for the five math tests. What score must he get on the fifth test in order to receive the highest grade?
Mathematics
2 answers:
castortr0y [4]3 years ago
5 0

Answer:98

Step-by-step explanation:

has to be higher then 97 so it has to be 98

Varvara68 [4.7K]3 years ago
4 0

He has to get 99% on his test to receive the highest grade in class

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If you buy 3 apples worth 0,65$ each with a 10$ note, how much change are you left with ?
Alex
$10.00 - 3($0.65)
$10.00 - $1.95
= $8.05
3 0
3 years ago
16x 2 + 8x - 15 Mathematics?
Gala2k [10]
If you're factoring, your answer is

(4x - 3)(4x + 5)

You'll work it out this way:

16x^2 + 8x - 15

Since there's no GCF here, you multiply the first and third terms, giving you -240. Your next step is to find two numbers that have a sum of 8 and a product of -240. These numbers are 20 and -12. Plug those in and you've got:

16x^2 + 20x - 12x - 15

From here you divide it into two binomials. These are (16x^2 + 20x) and (-12x - 15).
When you take out the greatest common factors (4x and -3), they become:

4x(4x + 5) - 3(4x + 5)

Then you group what's outside of the parentheses together (4x - 3)
And bring what's inside of the parentheses down (4x + 5).
This brings your answer to

(4x - 3)(4x + 5)
5 0
3 years ago
Solve for the variable.<br> The quotient of a number n divided by -4.5 equals 200.6.
jeyben [28]

Answer:

n = -902.7

Step-by-step explanation:

n/-4.5 = 200.6

n = (-4.5)(200.6)

n = -902.7

8 0
2 years ago
Determine whether the set of vectors <img src="https://tex.z-dn.net/?f=%20v_%7B1%3D%283%2C2%2C1%29%2C%20v_%7B2%7D%20%3D%28-1%2C-
Korolek [52]
Since each vector is a member of \mathbb R^3, the vectors will span \mathbb R^3 if they form a basis for \mathbb R^3, which requires that they be linearly independent of one another.

To show this, you have to establish that the only linear combination of the three vectors c_1\mathbf v_1+c_2\mathbf v_2+c_3\mathbf v_3 that gives the zero vector \mathbf0 occurs for scalars c_1=c_2=c_3=0.

c_1\begin{bmatrix}3\\2\\1\end{bmatrix}+c_2\begin{bmatrix}-1\\-2\\-4\end{bmatrix}+c_3\begin{bmatrix}1\\1\\-1\end{bmatrix}=(0,0,0)\iff\begin{bmatrix}3&-1&1\\2&-2&1\\1&-4&-1\end{bmatrix}\begin{bmatrix}c_1\\c_2\\c_3\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}

Solving this, you'll find that c_1=c_2=c_3=0, so the vectors are indeed linearly independent, thus forming a basis for \mathbb R^3 and therefore they must span \mathbb R^3.
4 0
3 years ago
Someone help! <br> Graph the solution to the inequality on the number line.<br> h &gt; 75.5
bazaltina [42]

Answer:

<h2>D</h2>

Step-by-step explanation:

<h2>hope it helps</h2>

....... .

8 0
2 years ago
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