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3 years ago

WILL GIVE BRAINLIEST IF RIGHT

Mathematics

2 answers:

Sindrei [870]3 years ago

8 0

Answer:

I think c

Step-by-step explanation:

When y=1 x=12

Sliva [168]3 years ago

4 0

Answer: The unit rate is 1/12

Step-by-step explanation:

You might be interested in

If three pears and four oranges cost $4.85 and three pears and ten oranges cost $8.75 what is the cost of a pear and an orange

marishachu [46]

Answer:

1 pear = $0.75; 1 orange = $0.65

Step-by-step explanation:

(1) 3P + 4O = 4.85

(2) 3P + 10O = 8.75 Eqn (2) - Eqn (1)

3P + 10O – 3P – 4O = 8.75 – 4.85 Combine like terms

6O = 3.90 Divide each side by 6

O = $0.65 Substitute into Eqn (1)

3P + 4×0.65 = 4.85

3P + 2.60 = 4.85 Subtract 2.60 from each side

3P = 2.25 Divide each side by 3

P = $0.75

Oranges cost $0.65 each and pears are $0.75 each

4 0

3 years ago

Please help 50 points!<br> Simplify the expressed

LenKa [72]

Answer:

Step-by-step explanation:

$\frac{x^{2} -x-12}{x+4} * \frac{2x^{2}-32 }{x^{2} -8x+16}$ rewrite as multiplication

$\frac{x^{2} +3x-4x-12}{x+4} * \frac{2(x^{2}-16) }{x^{2} -8x+16}$ look to factor

$\frac{x(x+3)-4(x+3)}{x+4} * \frac{2(x-4)(x+4) }{(x-4)^{2}}$ , difference of 2 squares and perfect square

$\frac{(x+3)(x-4)}{x+4} * \frac{2(x-4)(x+4) }{(x-4)^{2}}$ , finish factoring and simplify

2(x+3)

3 0

2 years ago

Linda commutes a total of 54 miles to and from work every day. She needs a new car, and good gas mileage is important to her. Sh

joja [24]

Answer:

she would save 1.1 extra gallons

Step-by-step explanation:

54 miles to work

30 miles a gallon so 1.8 gallons to get to work

27 miles a gallon so 2 gallons to get to work

30 miles car saves 5.4 miles more each day she drives to work

5.4 X 5 = 27 extra miles so 1.1 gallons she would be saving each work week

5 0

2 years ago

Mary Sweet works in a pizza store where busiest days fall on the weekend.

aliya0001 [1]

Answer:

$10.75 x 3 1/2 = $37.625 (rounded to the nearest cent: $37.63). $10.75 x 5 = $53.75. $37.625 + $53.75 = $91.375. (rounded to the nearest cent: $91.38) hope this helps

Step-by-step explanation:

- Zombie

4 0

2 years ago

Solve the following equations by factorisation method.Only factorisation not dharacharya.

Luba_88 [7]

Hello, please consider the following.

When $x_1$ and $x_2$ are two roots, we can factorise as

$ax^2+bx+c=a(x-x_1)(x-x_2)=a(x^2-(x_1+x_2)x+x_1x_2)=0$

So for the first equation, we can say that the sum of the zeros is

$\dfrac{a^2+b^2}{2}=\dfrac{a^2}{2}+\dfrac{b^2}{2}$

and the product is

$\dfrac{a^2b^2}{4}=\dfrac{a^2}{2}\dfrac{b^2}{2}$

So we can factorise as below.

$4x^2-2(a^2+b^2)x+a^2b^2=(2x-a^2)(2x-b^2)=0$

And the solutions are

$\boxed{\sf \n\bf \ \dfrac{a^2}{2} \ \ and \ \ \dfrac{b^2}{2}}$

For the second equation, we will complete the square and put the constant on the right side and take the root.

Let's do it!

$9x^2-9(a+b)x+2a^2+5ab+2b^2=0\\\\9(x-\dfrac{a+b}{2})^2-9\dfrac{(a+b)^2}{4}+2a^2+5ab+2b^2=0\\\\9(x-\dfrac{a+b}{2})^2=\dfrac{9(a+b)^2-8a^2-20ab-8b^2}{4}\\\\9(x-\dfrac{a+b}{2})^2=\dfrac{9a^2+18ab+9b^2-8a^2-20ab-8b^2}{4}\\\\9(x-\dfrac{a+b}{2})^2=\dfrac{a^2+b^2-2ab}{4}=\dfrac{(a-b)^2}{4}\\\\(x-\dfrac{a+b}{2})^2=\dfrac{(a-b)^2}{2^2\cdot 3^2}$

We take the root, and we find the two solutions

$\begin{aligned}x_1&=\dfrac{a+b}{2}-\dfrac{a-b}{6}\\\\&=\dfrac{3a+3b-a+b}{6}\\\\&=\dfrac{2a+4b}{6}\\\\&\boxed{=\dfrac{a+2b}{3}}\end{aligned}$

$\begin{aligned}x_1&=\dfrac{a+b}{2}+\dfrac{a-b}{6}\\\\&=\dfrac{3a+3b+a-b}{6}\\\\&=\dfrac{4a+2b}{6}\\\\&\boxed{=\dfrac{2a+b}{3}}\end{aligned}$

Thank you.

6 0

3 years ago