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3 years ago

I need help with #1 and 2

Mathematics

1 answer:

ollegr [7]3 years ago

8 0

Answer:

1) c. 11

Step-by-step explanation:

2) b. 4x² + 4

I hope these are correct its been a while since ive done these.

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Pls help!! I need the answer immediately

Strike441 [17]

Answer:

i). x³ + 9x² + yz - 15

ii). -21m³np - 8p⁵q + mnp + 4mn + 100

Step-by-step explanation:

Question (38)

i). Two expressions are -5x² - 4yz + 15 and x³+ 4x²- 3yz

By subtracting expression (1) from expression (2) we can the expression by addition which we can get expression (1).

(x³+ 4x²- 3yz) - (-5x² - 4yz + 15) = x³ + 4x² - 3yz + 5x² + 4yz - 15

= x³ + 9x² + yz - 15

ii). -15m³np + 2p⁵q - 6m³pn + mnp + 4mn - 10qp⁵+ 100

= (-15m³np - 6m³np) + (2p⁵q - 10qp⁵) + mnp + 4mn + 100

= -21m³np - 8p⁵q + mnp + 4mn + 100

4 0

4 years ago

The director of an animal shelter need to raise more than $50,000 during a fundraiser. Write an inequality that represents the a

mr_godi [17]

Answer:

m > 50,000

This means the money, m, will be greater than 50,000

5 0

3 years ago

Car lot one has a car to truck ratio of 3:4. Car lot two has a car to truck ratio of 2:3. Which car lot has more cars when both

Pani-rosa [81]

Answer:

this means that lot 1 would have more cars. there were 3 cars in lot 1 and 2 in lot 2

Step-by-step explanation:

lot 1- lot 2-

3:4 2:3

if both have 12 trucks, you need to multiply the number of cars with the number you used to make the trucks 12

3:4 would turn into 9:12

2:3 would be 8:12

this means that lot 1 would have more cars. there were 3 cars in lot 1 and 2 in lot 2

4 0

3 years ago

The high school band sells cured hams for a fundraiser. The first week of the

tigry1 [53]

Answer:

ITS THE SECOND ONE

Step-by-step explanation:

8 0

3 years ago

Compute the surface area of the portion of the sphere with center the origin and radius 4 that lies inside the cylinder x^2+y^2=

Tom [10]

Answer:

16π

Step-by-step explanation:

Given that:

The sphere of the radius = $x^2 + y^2 +z^2 = 4^2$

$z^2 = 16 -x^2 -y^2$

$z = \sqrt{16-x^2-y^2}$

The partial derivatives of $Z_x = \dfrac{-2x}{2 \sqrt{16-x^2 -y^2}}$

$Z_x = \dfrac{-x}{\sqrt{16-x^2 -y^2}}$

Similarly;

$Z_y = \dfrac{-y}{\sqrt{16-x^2 -y^2}}$

∴

$dS = \sqrt{1 + Z_x^2 +Z_y^2} \ \ . dA$

$=\sqrt{1 + \dfrac{x^2}{16-x^2-y^2} + \dfrac{y^2}{16-x^2-y^2} }\ \ .dA$

$=\sqrt{ \dfrac{16}{16-x^2-y^2} }\ \ .dA$

$=\dfrac{4}{\sqrt{ 16-x^2-y^2} }\ \ .dA$

Now; the region R = x² + y² = 12

Let;

x = rcosθ = x; x varies from 0 to 2π

y = rsinθ = y; y varies from 0 to $\sqrt{12}$

dA = rdrdθ

∴

The surface area $S = \int \limits _R \int \ dS$

$= \int \limits _R\int \ \dfrac{4}{\sqrt{ 16-x^2 -y^2} } \ dA$

$= \int \limits^{2 \pi}_{0} } \int^{\sqrt{12}}_{0} \dfrac{4}{\sqrt{16-r^2}} \ \ rdrd \theta$

$= 2 \pi \int^{\sqrt{12}}_{0} \ \dfrac{4r}{\sqrt{16-r^2}}\ dr$

$= 2 \pi \times 4 \Bigg [ \dfrac{\sqrt{16-r^2}}{\dfrac{1}{2}(-2)} \Bigg]^{\sqrt{12}}_{0}$

$= 8\pi ( - \sqrt{4} + \sqrt{16})$

= 8π ( -2 + 4)

= 8π(2)

= 16π

4 0

3 years ago