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2 years ago

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Answer the questions below

2 answers:

Leno4ka [110]2 years ago

5 0

I think this is what you're looking if not my apologies!

sergij07 [2.7K]2 years ago

5 0

Yea I think Jose is right

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. Question 1.1. What is the first step in solving x2 + 2x + 3 = 0 by completing the square?. (Points : 1). Factor the left side.

Assoli18 [71]

U r going to subtract 3 from each side of the equation....leaving you with :
x^2 + 2x = -3

6 0

3 years ago

Read 2 more answers

There are 765 marbles in a box. There are twice as many blue marbles as green marbles. There are three times as many red marbles

Digiron [165]

B=blue
g=green
r=red
2 times as many blue as green
b=2g
3 times as many red as blue
r=3b
subsitute 2g for b
r=3(2g)=6g
r=6g

r+g+b=765
so we subsitute
r=6g
g=g
b=2g

6g+g+2g=765
add like terms
9g=765
divide both sides by 9
g=85
subsitute
b=2b
b=2(85)
b=170

170 blue

5 0

3 years ago

I need help with this please

chubhunter [2.5K]

Answer:

1000L

Step-by-step explanation:

3000l=1kl

3000l×3kl

8 0

1 year ago

Evaluate the double integral. 2y2 dA, D is the triangular region with vertices (0, 1), (1, 2), (4, 1) D

zubka84 [21]

Answer:

$\mathbf{\iint _D y^2 dA= \dfrac{22}{3}}$

Step-by-step explanation:

From the image attached below;

We need to calculate the limits of x and y to find the double integral

We will notice that y varies from 1 to 2

The line equation for (0,1),(1,2) is:

$y-1 = \dfrac{2-1}{1-0}(x-0)$

y - 1 = x

The line equtaion for (1,2),(4,1) is:

$y-2 = \dfrac{1-2}{4-1}(x-1) \\ \\ y-2 = -\dfrac{1}{3}(x-1)$

-3(y-2) = (x -1)

-3y + 6 = x - 1

-x = 3y - 6 - 1

-x = 3y - 7

x = -3y + 7

This implies that x varies from y - 1 to -3y + 7

Now, the region D = {(x,y) | 1 ≤ y ≤ 2, y - 1 ≤ x ≤ -3y + 7}

The double integral can now be calculated as:

$\iint _D y^2 dA= \int ^2_1 \int ^{-3y +7}_{y-1} \ 2y ^2 \ dx \ dy$

$\iint _D y^2 dA= \int ^2_1 \bigg[ 2xy ^2 \bigg]^{-3y+7}_{y-1} \ dy$

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

$\iint _D y^2 dA= \int ^2_1 \bigg[-6y^3 +14y^2 -2y^3 +2y^2 \bigg ] \ dy$

$\iint _D y^2 dA= \int ^2_1 \bigg[-8y^3 +16y^2 \bigg ] \ dy$

$\iint _D y^2 dA= \bigg[-8(\dfrac{y^4}{4}) +16(\dfrac{y^3}{3})\bigg ] ^2_1$

$\iint _D y^2 dA= \bigg[-8(\dfrac{16}{4}-\dfrac{1}{4}) +16(\dfrac{8}{3}-\dfrac{1}{3})\bigg ]$

$\iint _D y^2 dA= \bigg[-8(\dfrac{15}{4}) +16(\dfrac{7}{3})\bigg ]$

$\iint _D y^2 dA= -30 + \dfrac{112}{3}$

$\iint _D y^2 dA= \dfrac{-90+112}{3}$

$\mathbf{\iint _D y^2 dA= \dfrac{22}{3}}$

4 0

3 years ago

Explain how to find 80% of a number if 135% of that number is 540.

Dmitry_Shevchenko [17]

Answer:

x = 911.25

Step-by-step explanation:

let the number be x

by putting the word problem in equation, we get :

80/100 × x = 135/100 ×540
4/5 × x = 135/10 × 54
x = 135/10 × 54 × 5/4
x = 911.25

6 0

2 years ago