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Mars2501 [29]
3 years ago
12

Given F ( x ) = -2/3 x - 4 What is the zero of this function?

Mathematics
1 answer:
sineoko [7]3 years ago
5 0

Answer:

-4

Step-by-step explanation:

f(0)=(-2/3)(0) - 4

= - 4

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What is the answer now in two minutes
Wittaler [7]

Answer:

60 degrees

Step-by-step explanation:

Cos of 60 is 1/2, unit circles helps with this.

4 0
3 years ago
The area of the triangle above will equal one half of a rectangle that is 5 units long and units wide. (Input only whole numbers
il63 [147K]

Answer:

<h2>Width of the rectangle is 2</h2>

Step-by-step explanation:

A rectangle can is made up of two right angled triangle as shown.

Since area of a rectangle = base * height

Area of one of the triangle = \frac{1}{2}\  * \ area\ of\ the\ rectangle

The length and width of the triangle will be the same as that of the rectangle i.e  5units long and 2units wide

Area of one of the triangle = \frac{1}{2}  * 5 * 2

Area of one of the triangle = 5units²

The area of the triangle above will equal one half of a rectangle that is 5 units long and 2 units wide.

5 0
3 years ago
Can somebody plz help answer these questions correctly thanks! (Grade7math) btw :)
Monica [59]

Answer:

1. both boxes 3

2. both boxes 7

3. both boxes 5

4. both boxes 2

5. top box 12 bottom 18

6. top box 9 bottom box 11

7. top box 42 bottom box 60

8. top box 4 bottom box 5

7 0
3 years ago
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
3 years ago
2 thousands divided by what number is equal to 2 hundreds?
laila [671]

Answer:

10

Step-by-step explanation:

it is 10 2000÷10=200

8 0
3 years ago
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