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3 years ago

Please help With the following question

Mathematics

1 answer:

Vikki [24]3 years ago

4 0

The equation youre using is y=mx+b

b is the y int which 3 and the slope between the points is 5/2 so your equation is y=5/2x+3

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Solve the system of equations using the linear combination method {5p-3q=-39 -2p-3q=3

VikaD [51]

Answer:

p=-6

q=3

Step-by-step explanation:

5p-3q=-39

-2p-3q=3

Multiply the second equation by -1

2p +3q = -3

Add the first equation and the modified second equation

5p-3q=-39

2p +3q = -3

---------------------

7p = -42

Divide by 7

7p/7 = -42/7

p = -6

Now we can find q

2p +3q = -3

2(-6) +3q = -3

-12 +3q = -3

Add 12 to each side

-12+12 +3q = -3+12

3q = 9

Divide by 3

3q/3 = 9/3

q=3

6 0

3 years ago

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There is 6/8 of a cake leftover after a birthday party. How many 1/4 pieces can be made out of the leftover cake?

ikadub [295]

3/4 because if you make it smaller you would get that

4 0

3 years ago

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(24x3 − 14x2 + 20x + 6) ÷ (4x2 − 3x + 5) = Q + <br> R<br> 4x2 − 3x + 5

riadik2000 [5.3K]

Answer:

Q= 6x+1

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Step-by-step explanation:

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3 years ago

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Mr. King has 36 pair of shoes. 3/4 of them are gym shoes, how many of his shoes are gym shoes

Alex777 [14]

Answer:

The number of pair of gym shoes does Mr. king has is 27 pairs .

Step-by-step explanation:

Given as :

Total number of pairs of shoes does king has = 36 pairs

The number of gym shoes = $\dfrac{3}{4}$ of the total pairs of shoes

Let The number of pair of gym shoes = n

<u>According to question</u>

The number of gym shoes = $\dfrac{3}{4}$ × the total pairs of shoes

Or, n = $\dfrac{3}{4}$ × 36

i.e n = $\frac{3\times 36}{4}$

Or, n = 3 × 9

∴ n = 27

So, The number of pair of gym shoes = n = 27 pairs

Hence, The number of pair of gym shoes does Mr. king has is 27 pairs . Answer

6 0

3 years ago

P(x)= 3x^3-5x^2-14x-4

nexus9112 [7]

$\displaystyle\\
P(x)=3x^3-5x^2-14x-4\\\\
D_{-4}=\{-4;~-2;~\underline{\bf -1};~1;~2;~4\}\\\\
\text{We observe that } \frac{-1}{3} \text{ is a solution of the equation:}\\
3x^3-5x^2-14x-4=0\\\\
$

$\displaystyle\\
\text{Verification}\\\\
3x^3-5x^2-14x-4=\\\\
=3\times\Big(-\frac{1}{3}\Big)^3-5\times\Big(-\frac{1}{3}\Big)^2-14\times\Big(-\frac{1}{3}\Big)-4=\\\\
=-\frac{1}{9}-\frac{5}{9}+\frac{14}{3}-4=\\\\
=-\frac{6}{9}+\frac{14}{3}-4=\\\\
=-\frac{6}{9}+\frac{42}{9}- \frac{4\times 9}{9}=\\\\
 =-\frac{6}{9}+\frac{42}{9}- \frac{36}{9}= \frac{42-6-36}{9}=\frac{42-42}{9}=\frac{0}{9}=0\\\\
\Longrightarrow~~~P(x)~\vdots~\Big(x+ \frac{1}{3}\Big)\\\\
\Longrightarrow~~~P(x)~\vdots~(3x+1)$

$\displaystyle\\
3x^3-5x^2-14x-4=0\\
~~~~~-5x^2 = x^2 - 6x^2\\
~~~~~-14x =-2x-12x \\
3x^3+x^2 - 6x^2-2x-12x-4=0\\
x^2(3x+1)-2x(3x+1) -4(3x+1)=0\\
(3x+1)(x^2-2x -4)=0\\\\
\text{Solve: } x^2-2x -4=0\\\\
x_{12}= \frac{-b\pm \sqrt{b^2-4ac}}{2a}=\\\\=\frac{2\pm \sqrt{4+16}}{2}=\frac{2\pm \sqrt{20}}{2}=\frac{2\pm 2\sqrt{5}}{2}=1\pm\sqrt{5}\\\\
x_1 =1+\sqrt{5}\\
x_2 =1-\sqrt{5}\\
\Longrightarrow P(x)= 3x^3-5x^2-14x-4 =\boxed{(3x+1)(x-1-\sqrt{5})(x-1+\sqrt{5})}$

7 0

3 years ago