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2 years ago

Please help, I'm so lost. :(

Mathematics

2 answers:

steposvetlana [31]2 years ago

7 0

The answer is g= 49
Explanation)
Remove the radical by raising each side to the index of the radical

Lunna [17]2 years ago

5 0

Answer:

g = 35^5 = 52,521,875

Step-by-step explanation:

g = 35^5

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The question for this problem is “The owner of a food cart sells an average of 120 frozen treats per day during the summer”.

ziro4ka [17]

Answer:

i think its b

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Log 15 (2x − 2) = log 15 (x+9)

notka56 [123]

Answer:

$x=11$

Step-by-step explanation:

Given the equation

$\log_{15}(2x-2)=\log_{15}(x+9)$

Note that

$2x-2>0\\ \\2x>2\\ \\x>1$

and

$x+9>0\\ \\x>-9$

Combined, $x>1$

Solve the equation:

$2x-2= x+9\\ \\2x-x=9+2\\ \\x=11$

Since $11>1,$ this is the solution to the equation.

5 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

The area of a square bedroom is 144 square feet. what is the length of one wall?

Rufina [12.5K]

The answer is: "12 feet" .
________________________________
Note: In a square, the length of EACH of the four sides of the square is the same.

Area = Length * width.

For a square, length = width.

So for a square, Area = length * width = (length of a side)² = s² ,

Given: A = s² = 144 ft² ;

Solve for the positive value of "s" .
________________________________
  → s² = 144 ft² ; Take the "square root" of each side ;

  → √(s²) = √(144 ft²) ;

  → s = 12 ft.
_______________________________
The answer is: 12 ft.
_______________________________

6 0

3 years ago

Lorico [155]

Answer:

Kindly check explanation

Step-by-step explanation:

Given the following :

Price of home = $220,000

Down payment = 20%

3 points at closing

30 years fixed rate mortgage at 7%

A.)The down payment :

20% of price

0.2 × $220,000 = $44,000

B.) Amount of the Mortgage :

Price - down payment

$220,000 - $44,000 = $176,000

C.) Amount to be paid for the 3 point of closing :

3% of mortgage amount

0.03 * $176,000 = $5,280

D.) monthly payment :

(mortgage amount(1 + rate)) / 30* 12

176000(1 + 0.07)

($176,000 + $12320) / 360

$523

E.) Total cost of interest :

0.07 * 176000 = $12,320

8 0

3 years ago