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3 years ago

30 Points ill mark brainiest, please help.

Mathematics

2 answers:

meriva3 years ago

6 0

Answer:

a = 1, b = -1, c = -2

Or,

a = -1, b = 1, c = 2

Step-by-step explanation:

–2 = –x + x² – 4

x² - x - 4 + 2 = 0

x² - x - 2 = 0

a = 1, b = -1, c = -2

Or,

–2 = –x + x² – 4

-x² + x - 2 + 4 = 0

-x² + x + 2 = 0

a = -1, b = 1, c = 2

melomori [17]3 years ago

5 0

Answer:

Got it!

The given Quadratic Equation is:

-2= -x+x^2-4

i.e, x^2-x-2=0

now we have to compare it to the form ax^2+bx+c=0

by comparing we got a=1,b= -1 and c= -2

in other way around ,when coefficient of x^2 taken as negative then the sign of a,b&c will change ,

as a= -1, b=1 and c=2

✌️:)

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A mother invests $7000 in a bank account at the time of her daughter's birth. The interest is 18) compounded quarterly at a rate

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1 year ago

The two dot plots show the number of miles ran by 14 students at the beginning and at the end of the school year compare each me

Naddika [18.5K]

The two dot plots are missing, so i have attached it.

Answer:

The mean at the beginning of the school year was 9.5 miles and the mean at the end of the school year was 10.2 miles

Step-by-step explanation:

From the attached image, we are told to compare the means for each plot to the nearest tenth.

Mean = Σx/n

Now, from the image, total number of miles run by the 14 students at the beginning of the school year is;

(1 × 7) + (2 × 8) + (4 × 9) + (4 × 10) + (2 × 11) + (1 × 12) = 133

Mean of miles run at the beginning of the school year = 133/14 = 9.5 miles

Again, from the table, total miles run at the end of the school year = (2 × 8) + (2 × 9) + (4 × 10) + (3 × 11) + (3 × 12) = 143

Mean of miles run at the end of the school year = 143/14 = 10.2 miles

Thus;

The mean at the beginning of the school year was 9.5 miles and the mean at the end of the school year was 10.2 miles

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3 years ago

Michael’s weight can be represented by the expression 72x^5. Al’s weight can be represented by the expression 9x^7.

podryga [215]

Answer:

a) $\frac{72x^5}{9x^7}$

b) $=\frac{8}{x^2}$

c), Yes; $8x^2$

Step-by-step explanation:

Michael's Weight: $72x^5$

Al's weight: $9x^7$

a) Ratio of Michael's weight to Al's weight: $\frac{72x^5}{9x^7}$

b) This ratio simplifies to: $\frac{8\times 9x^5}{9x^5\timesx^2}$

$=\frac{8}{x^2}$

c) Yes, If the exponent in each expression were negative, then we have:

Ratio of Michael's weight to Al's weight: $\frac{72x^{-5}}{9x^{-7}}$

This ratio simplifies to: $8x^{-5--7}=8x^2$

The two ratios are not the same.

3 0

3 years ago

Write an equation in slope-intercept form for the line that is parallel to y=−4x−3 and that passes through the point (−2,4).

kirill115 [55]

Answer:

The Correct option is Last one $y=-4x-4$

Therefore, equation of the line in slope-intercept that passes through (-2,4) and is parallel to the line $y=-4x-3$ is $y=-4x-4$

Step-by-step explanation:

Given:

$y=-4x-3$

To Find:

Equation of line passing through ( -2, 4) and is parallel to the line y=-4x-3

Solution:

$y=-4x-3$ ..........Given

Comparing with Slope-Intercept form,

$y=mx+c$

Where m =slope

We get

$Slope = m = -4$

We know that parallel lines have Equal slopes.

Therefore the slope of the required line passing through (-2 , 4) will also have the slope = m = -4.

Now the equation of line in slope point form given by

$(y-y_{1})=m(x-x_{1})$

Substituting the points and so we will get the required equation of the line,

$(y-4))=-4(x-(-2))=-4x-8\\\\y=-4x-8+4=-4x-4\\\\y=-4x-4......Equation\ of\ line$

Therefore, equation of the line in slope-intercept that passes through (-2,4) and is parallel to the line $y=-4x-3$ is $y=-4x-4$

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3 years ago

Y = x + 2<br> 2x - y = -4<br> Solve the system of equations using substitution.

const2013 [10]

The answer is (-2,0)

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3 years ago