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3 years ago

Please help! Will give brainly, 50 points!! I'm stuck with this question and I don't get it!

Mathematics

2 answers:

pishuonlain [190]3 years ago

7 0

Answer: The answer is x^2 + 4x - 3

Step-by-step explanation:

You just combine like terms, the only ones that can combine is the 6x and -4x. You just subtract them to get the answer of x^2 + 4x - 3.

Pachacha [2.7K]3 years ago

7 0

Answer:

h(x) + g(x) = x² + 6x - 2x - 3

h(x) + g(x) = x² + 4x - 3

Simply add the like terms and leave the other terms

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3 years ago

A bakery sold 208 chocolate cupcakes in a day, which was 52% of the total number of cupcakes sold that day. How many total cupca

Dima020 [189]

Answer:

Step-by-step explanation:

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3 years ago

Drag each expression to show whether it can be used to find the volume, surface area, or neither.

Mumz [18]

Answer:

Volume = 12 *6 *8 , Surface area = 2 ( 12 *6 + 8 *12 + 6* 8 )

Step-by-step explanation:

Given : Cuboid with length 12 , width 6 and height 8 units.

To find : Drag each expression to show whether it can be used to find the volume, surface area, or neither.

Solution : We have given Cuboid with

Length = 12 units ,

Width = 6 units

Height = 8 units.

Volume of cuboid = length * width * height .

Volume = 12 *6 *8.

Surface area = 2 ( l *w + h *+w *h)

Surface area = 2 ( 12 *6 + 8 *12 + 6* 8 ).

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7 0

3 years ago

Read 2 more answers

Evaluate the double integral.

Fynjy0 [20]

Answer:

$\iint_D 8y^2 \ dA = \dfrac{88}{3}$

Step-by-step explanation:

The equation of the line through the point $(x_o,y_o)$ & $(x_1,y_1)$ can be represented by:

$y-y_o = m(x - x_o)$

Making m the subject;

$m = \dfrac{y_1 - y_0}{x_1-x_0}$

∴

we need to carry out the equation of the line through (0,1) and (1,2)

i.e

y - 1 = m(x - 0)

y - 1 = mx

where;

$m= \dfrac{2-1}{1-0}$

m = 1

Thus;

y - 1 = (1)x

y - 1 = x ---- (1)

The equation of the line through (1,2) & (4,1) is:

y -2 = m (x - 1)

where;

$m = \dfrac{1-2}{4-1}$

$m = \dfrac{-1}{3}$

∴

$y-2 = -\dfrac{1}{3}(x-1)$

-3(y-2) = x - 1

-3y + 6 = x - 1

x = -3y + 7

Thus: for equation of two lines

x = y - 1

x = -3y + 7

i.e.

y - 1 = -3y + 7

y + 3y = 1 + 7

4y = 8

y = 2

Now, y ranges from 1 → 2 & x ranges from y - 1 to -3y + 7

∴

$\iint_D 8y^2 \ dA = \int^2_1 \int ^{-3y+7}_{y-1} \ 8y^2 \ dxdy$

$\iint_D 8y^2 \ dA =8 \int^2_1 \int ^{-3y+7}_{y-1} \ y^2 \ dxdy$

$\iint_D 8y^2 \ dA =8 \int^2_1 \bigg ( \int^{-3y+7}_{y-1} \ dx \bigg) dy$

$\iint_D 8y^2 \ dA =8 \int^2_1 \bigg ( [xy^2]^{-3y+7}_{y-1} \bigg ) \ dy$

$\iint_D 8y^2 \ dA =8 \int^2_1 \bigg ( [y^2(-3y+7-y+1)]\bigg ) \ dy$

$\iint_D 8y^2 \ dA =8 \int^2_1 \bigg ([y^2(-4y+8)] \bigg ) \ dy$

$\iint_D 8y^2 \ dA =8 \int^2_1 \bigg ( -4y^3+8y^2 \bigg ) \ dy$

$\iint_D 8y^2 \ dA =8 \bigg [\dfrac{ -4y^4}{4}+\dfrac{8y^3}{3} \bigg ]^2_1$

$\iint_D 8y^2 \ dA =8 \bigg [ -y^4+\dfrac{8y^3}{3} \bigg ]^2_1$

$\iint_D 8y^2 \ dA =8 \bigg [ -2^4+\dfrac{8(2)^3}{3} + 1^4- \dfrac{8\times (1)^3}{3}\bigg]$

$\iint_D 8y^2 \ dA =8 \bigg [ -16+\dfrac{64}{3} + 1- \dfrac{8}{3}\bigg]$

$\iint_D 8y^2 \ dA =8 \bigg [ -15+ \dfrac{64-8}{3}\bigg]$

$\iint_D 8y^2 \ dA =8 \bigg [ -15+ \dfrac{56}{3}\bigg]$

$\iint_D 8y^2 \ dA =8 \bigg [ \dfrac{-45+56}{3}\bigg]$

$\iint_D 8y^2 \ dA =8 \bigg [ \dfrac{11}{3}\bigg]$

$\iint_D 8y^2 \ dA = \dfrac{88}{3}$

4 0

2 years ago

Members of the student concil are conducting a fundraiser by selling school calendars. After selling 80 calendars, they had a lo

Oksana_A [137]

Answer:

a) $y=8x-1000$

b) $profit=\$8$

c) They'd have lost $1000 if they had sold no calendars.

Step-by-step explanation:

a) The equation of the line in Slope-Intercept form is:

$y=mx+b$

Where "m" is the slope and "b" is the y-intercept.

In this case we know that "y" represents the profit of loss and "x" the number of calendars sold.

Then, according to the exercise, the line passes through these two points:

$(80,-360)$ and $(200,600)$

Then, we can find the slope of the line with the formula $m=\frac{y_2-y_1}{x_2-x_1}$

$m=\frac{-360-600}{80-200}=8$

Now, we can substitute the slope and one of those points into $y=mx+b$ and solve for "b":

$600=8(200)+b\\\\b=-1000$

Then, subtituting values, we get that the equation that describes the relation between the profit of loss and the number of calendars sold, is:

$y=8x-1000$

b) The slope of the line is the profit they made from selling each calendar

$profit=8$

c) The y-intercept is the amount they would have lost if they had sold no calendars:

$b=-1000$

They'd have lost $1000 if they had sold no calendars.

6 0

2 years ago