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4 years ago

How do u solve this?

Mathematics

1 answer:

Ivenika [448]4 years ago

5 0

Answer:

Step-by-step explanation:

Tuesday : -1/2

Wednesday + 3/4

Thursday : -3/8

Add them together

-1/2 + 3/4- 3/8

Get a common denominator

-4/8 + 6/8 - 3/8

-1/8

The closest integer value to -1/8 is 0

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Derrick really wants to earn at least $330 at the Bike Shop this week before he goes on vacation. He earns $150 in a week plus $

777dan777 [17]

Answer:

150 + 20x ≥ 330

Step-by-step explanation:

"At least" indicates that you should use the greater than or equal to sign. His weekly wage and the bicycle assembly money when added must be at least 330, so 150 and 20x are on the same side. The 20x represents the $20 for each bicycle multiplied by the amount of bicycles he assembled.

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3 years ago

How who I complete the table?

Natalka [10]

Answer:

Step-by-step explanation:

you have to remove // to calculate

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4 years ago

Owen’s parents were planning a birthday party for him. They have $20.00 to spend on goodie bags. They need to buy 6 goodie bags

Dimas [21]

Answer:

Yes, they can buy all the bags.

Step-by-step explanation:

Owen's parents need to buy 6 bags for $2.49 each.

So, we need to calculate the total price:

$2.49 * 6 = $14.94

Since the bags cost $14.94, and Owen's parents have $20.00, they have enough money to buy all 6 bags. 14.94 < 20.00

Hope this helps!

5 0

3 years ago

Read 2 more answers

What is the quotient of 5.3t -(20÷4)+ 11

lana66690 [7]

Answer:

-9.7t

Step-by-step explanation:

PEMDAS:

Parenthesis: (20 / 4) = 5

Exponents

Multiply

Divide

Add: 5 + 11 = 15

Subtract: 5.3t - 15 = <u>-9.7t</u>

6 0

3 years ago

Use the given transformation x=4u, y=3v to evaluate the integral. ∬r4x2 da, where r is the region bounded by the ellipse x216 y2

exis [7]

The Jacobian for this transformation is

$J = \begin{bmatrix} x_u & x_v \\ y_u & y_v \end{bmatrix} = \begin{bmatrix} 4 & 0 \\ 0 & 3 \end{bmatrix}$

with determinant $|J| = 12$ , hence the area element becomes

$dA = dx\,dy = 12 \, du\,dv$

Then the integral becomes

$\displaystyle \iint_{R'} 4x^2 \, dA = 768 \iint_R u^2 \, du \, dv$

where $R'$ is the unit circle,

$\dfrac{x^2}{16} + \dfrac{y^2}9 = \dfrac{(4u^2)}{16} + \dfrac{(3v)^2}9 = u^2 + v^2 = 1$

so that

$\displaystyle 768 \iint_R u^2 \, du \, dv = 768 \int_{-1}^1 \int_{-\sqrt{1-v^2}}^{\sqrt{1-v^2}} u^2 \, du \, dv$

Now you could evaluate the integral as-is, but it's really much easier to do if we convert to polar coordinates.

$\begin{cases} u = r\cos(\theta) \\ v = r\sin(\theta) \\ u^2+v^2 = r^2\\ du\,dv = r\,dr\,d\theta\end{cases}$

Then

$\displaystyle 768 \int_{-1}^1 \int_{-\sqrt{1-v^2}}^{\sqrt{1-v^2}} u^2\,du\,dv = 768 \int_0^{2\pi} \int_0^1 (r\cos(\theta))^2 r\,dr\,d\theta \\\\ ~~~~~~~~~~~~ = 768 \left(\int_0^{2\pi} \cos^2(\theta)\,d\theta\right) \left(\int_0^1 r^3\,dr\right) = \boxed{192\pi}$

3 0

2 years ago