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2 years ago

Find the slope of the graph of C

Mathematics

1 answer:

alexandr402 [8]2 years ago

7 0

Step-by-step explanation:

the slope of a line or linear equation is always the factor of the driving variable.

usually in the form of

y = ax + b

x is the variable, and a the slope.

but in our case the variable is called "n".

it does not matter, the same principle applies.

the slope is the factor of n : 1050.

and as the slope is always the ratio y/x, we are looking for a fraction form, and that would be 1050/1.

so, the third answer is correct.

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3(v + 2) + 4 = 13<br> HELP ME!!

Zolol [24]

3(v + 2) + 4 = 13

Mutiply the bracket by 3

(3)(v)= 3v

(3)(2)= 6

3v+6+4= 13

3v+10=13

Move +10 to the other side

Sign changes from +10 to -10

3v+10-10= 13-10

3v= 13-10

3v= 3

divide both sides by 3

3v/3= 3/3

v= 1

Answer : v= 1

4 0

4 years ago

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What is 16+48 in distributive property??

Butoxors [25]

Answer:

un-distribute the GCF of the two numbers.

16(1 + 3)

Distributing we get:

16(1) + 16(3)

16 + 48

Step-by-step explanation:

7 0

3 years ago

No links or trolls please. I need help with this math, it’s due soon. Thanks you. It’s a division sign by the way, it’s hard to

xxTIMURxx [149]

Answer:

n^2p

Step-by-step explanation:

Thats the answer

3 0

3 years ago

. Use the quadratic formula to solve each quadratic real equation. Round

Liono4ka [1.6K]

Answer:

A. No real solution

B. 5 and -1.5

C. 5.5

Step-by-step explanation:

The quadratic formula is:

$\begin{array}{*{20}c} {\frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}}} \end{array}$ , with a being the x² term, b being the x term, and c being the constant.

Let's solve for a.

$\begin{array}{*{20}c} {\frac{{ 5 \pm \sqrt {5^2 - 4\cdot1\cdot11} }}{{2\cdot1}}} \end{array}$

$\begin{array}{*{20}c} {\frac{{ 5 \pm \sqrt {25 - 44} }}{{2}}} \end{array}$

$\begin{array}{*{20}c} {\frac{{ 5 \pm \sqrt {-19} }}{{2}}} \end{array}$

We can't take the square root of a negative number, so A has no real solution.

Let's do B now.

$\begin{array}{*{20}c} {\frac{{ 7 \pm \sqrt {7^2 - 4\cdot-2\cdot15} }}{{2\cdot-2}}} \end{array}$

$\begin{array}{*{20}c} {\frac{{ 7 \pm \sqrt {49 + 120} }}{{-4}}} \end{array}$

$\begin{array}{*{20}c} {\frac{{ 7 \pm \sqrt {169} }}{{-4}}} \end{array}$

$\begin{array}{*{20}c} {\frac{{ 7 \pm 13 }}{{-4}}} \end{array}$

$\frac{7+13}{4} = 5\\\frac{7-13}{4}=-1.5$

So B has two solutions of 5 and -1.5.

Now to C!

$\begin{array}{*{20}c} {\frac{{ -(-44) \pm \sqrt {-44^2 - 4\cdot4\cdot121} }}{{2\cdot4}}} \end{array}$

$\begin{array}{*{20}c} {\frac{{ 44 \pm \sqrt {1936 - 1936} }}{{8}}} \end{array}$

$\begin{array}{*{20}c} {\frac{{ 44 \pm 0}}{{8}}} \end{array}$

$\frac{44}{8} = 5.5$

So c has one solution: 5.5

Hope this helped (and I'm sorry I'm late!)

4 0

3 years ago

what side lengths should be used to model a rectangle with an area of x^2+3x-10 square units? x-5 and x+2 x and x+3 x+3 and x-10

klemol [59]

Answer:

Answer: x-2 and x+5 side lengths should be used to model a rectangle.

Step-by-step explanation:

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5 0

3 years ago

Read 2 more answers