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2 years ago

$25000 at 4.25% for 2 year

Mathematics

1 answer:

lianna [129]2 years ago

6 0

Answer:

Step-by-step

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Mark brainliest if this answer is correct please

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. Write the following expression in factored form using the greatest common factor.<br> 50a3 + 10a2

Tanya [424]

The factors of 50a³ are 1, 2, 5, 10, 25, 50,
   and their products with a, a² and a³ .

The factors of 10a² are 1, 2, 5, 10,
    and their products with 'a' and a² .

Their common factors are 1, 2, 5, 10,
   and their products with 'a' and a².

Their greatest common factor is 10a² .

(Another way to spot it, easily, is to remember this helpful factoid:

   If the smaller number is a factor of the larger number,
   then the smaller number is their greatest common factor.

Using the greatest common factor, then . . .

   50a³ + 10a² =   10a²(5a + 1) .

3 0

3 years ago

In the figure above, are <3 and <7 alternate interior angles,

icang [17]

Answer:

Corresponding angles

Step-by-step explanation:

<3 and <7 lie on the same side of the transversal line. Each angle also is located on the outside of each parallel lines cut across by the transversal.

Therefore, <3 and <7 are corresponding angles.

7 0

2 years ago

Find the domain and range of the function f(x) = x -|2 cos x|

adell [148]

Answer:

Step-by-step explanation:

3 0

2 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

2 years ago

CAN SOMEONE HELP ME UNDERSTAND THIS PLEASE

ICE Princess25 [194]

YEAH OFC SHAWTY!

so for A its -4^11 because if the powers have the same base and they are being multiplied you add the powers

For B its the same idea, they have the same base, and are being multiplied so you add the powers- 13^9

For C its similar, you have the same base, but since its Dividing, you subtract the powers. So, 9^5
for D you pretend like the denominator has a power of 1 and subtract. So its -24^5-24^1 which hopefully puts it into perspective of being -24^4 becuase we subtracted the powers

I cant do 2 right now, but i will in a minute

<u>And for 3 his mistake is that they arent the same bases so he cant add the powers, he should have converted them to the same base first. </u>

7 0

2 years ago