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Jobisdone [24]
4 years ago
8

What is the product? (6 r minus 1)(negative 8 r minus 3)

Mathematics
2 answers:
just olya [345]4 years ago
3 0

Answer:

The required product is -48r^2-10r+3

Step-by-step explanation:

Consider the provided expression.

(6r-1)(-8r-3)

Open the parentheses.

6r\left(-8r\right)+6r\left(-3\right)+\left(-1\right)\left(-8r\right)+\left(-1\right)\left(-3\right)

-6\cdot \:8rr-6\cdot \:3r+1\cdot \:8r+1\cdot \:3

-48r^2-18r+8r+3

Add the like terms.

-48r^2-10r+3

Hence, the required product is -48r^2-10r+3

Marta_Voda [28]4 years ago
3 0

Answer:

-47r²-10r+3

Step-by-step explanation:

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What the range of 80,84,86,87,90,91,92,94,100
inessss [21]

Answer:

20

pls mark brainliest if possible

Step-by-step explanation:

the range is the difference between the values of the largest and smallest values. You can figure this out with subtraction.

Largest number: 100

Smallest number: 80

Range: 20

6 0
3 years ago
How do solve this problem? <br> 1 2/5x =7
elena55 [62]

Answer:

if you are solving for x then the answer is 5

Step-by-step explanation:

5 0
3 years ago
In an apartment complex with 28 units, 19 of the renters keep a pet. What percentage does not keep a pet?
Aleks04 [339]

If 19 out of 28 renters keep a pet, there are 28-19 = 9 renters who don't keep a pet.

Whenever you have a subset of some set, and you want to know which percentage of the set the subset represents, you simply have to compute

\dfrac{\text{number of elements in the subset}}{\text{number of elements in the set}}\times 100

So, in your case, you're wondering what percentage of 28 does 9 represent. So, the formula becomes

\dfrac{9}{28}\times 100 = 0.32\overline{142857}\times 100 = 32.\overline{142857} \approx 32.14\%

6 0
3 years ago
What are the solutions of 8x2 = 6 + 22x? Check all of the boxes that apply.
Tamiku [17]

Answer:

x = 3 x = ‐1/4

Step-by-step explanation:

Was right on the quiz

6 0
3 years ago
Read 2 more answers
Let f ( ) sin(arctan ) x x = . What is the range of f ?
nikklg [1K]
<span>So you have composed two functions,
</span><span>h(x)=sin(x) and g(x)=arctan(x)</span>
<span>→f=h∘g</span><span>
meaning
</span><span>f(x)=h(g(x))</span>
<span>g:R→<span>[<span>−1;1</span>]</span></span>
<span>h:R→[−<span>π2</span>;<span>π2</span>]</span><span>
And since
</span><span>[−1;1]∈R→f is defined ∀x∈R</span><span>

And since arctan(x) is strictly increasing and continuous in [-1;1] ,
</span><span>h(g(]−∞;∞[))=h([−1;1])=[arctan(−1);arctan(1)]</span><span>
Meaning
</span><span>f:R→[arctan(−1);arctan(1)]=[−<span>π4</span>;<span>π4</span>]</span><span>
so there's your domain</span>
5 0
3 years ago
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