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vladimir1956 [14]

2 years ago

13

At the James Factory, the assembly line produces 89 pairs of jeans in a batch. The pairs of jeans are packaged in boxes with 11

each. If 7 batches are made in one day, how many pairs of jeans will be leftover at the end of it?

1 answer:

kobusy [5.1K]2 years ago

7 0

Answer:

7 pairs left over

Step-by-step explanation:

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8 friends share 7 pizzas. How much pizza does each person get? Give your answer as a decimal and as a fraction.

dmitriy555 [2]

Answer:

0.875 as a decimal
875/1000 or 7/8 as a fraction

explanation:

7/8 = 0.875
875/1000 = 0.875

hope this helps :)

4 0

2 years ago

Read 2 more answers

What is correct in these ordered pairs (4,9) and (4,-9)?

timurjin [86]

Answer:

(maaf kalau salah

THANKS

4 0

2 years ago

Trudy bought 1 foot of fabric. If she cuts the fabric into 3-inch pieces, how many pieces will she have?

WINSTONCH [101]

Answer:

Trudy will have 4 pieces of the fabric.

Step-by-step explanation:

Since, 1 foot of the fabric = 12 inches

If Trudy cuts the fabric into pieces of 3 inch.

Number of pieces she will have = $\frac{\text{Total length of the fabric}}{\text{Length of one piece}}$

= $\frac{12}{3}$

= 4 pieces

Therefore, she will have 4 pieces of the fabric.

5 0

2 years ago

NISA [10]

<h2>✒️Area Between Curves</h2>

$\small\begin{array}{ |c|c} \hline \bold{Area\ Between\ Curves} \\ \\ \textsf{Solving for the intersection of }\rm y = x^2 + 2\textsf{ and }\\ \rm y = 4, \\ \\ \qquad \begin{aligned} \rm y_1 &=\rm y_2 \\ \rm x^2 + 2 &=\rm 4 \\ \rm x^2 &= \rm 2 \\ \rm x &=\rm \pm \sqrt{2} \end{aligned} \\ \\ \textsf{We only need the first quadrant area bounded} \\ \textsf{by the given curves so the integral for the area} \\ \textsf{would then be} \\ \\ \boldsymbol{\displaystyle \rm A = \int_{\ a}^{\ b} {\left( \begin{array}{c}\text{upper} \\ \text{function}\end{array} \right) - \left( \begin{array}{c} \text{lower} \\ \text{function} \end{array} \right)\ dx}} \\ \\ \displaystyle \rm A = \int_{0}^{\sqrt{2}} \Big[4 - (x^2 + 2)\Big]\ dx \\ \\ \displaystyle \rm A = \int_{0}^{\sqrt{2}} (2 - x^2)\ dx \\ \\ \rm A = \left[2x - \dfrac{x^3}{3}\right]_{0}^{\sqrt{2}} \\ \\ \rm A = 2\sqrt{2} - \dfrac{\big(\sqrt{2}\big)^3}{3} \\ \\ \rm A = 2\sqrt{2} - \dfrac{2\sqrt{2}}{3} \\ \\\red{\boxed{\begin{array}{c} \rm A = \dfrac{4\sqrt{2}}{3}\textsf{ sq. units} \\ \textsf{or} \\ \rm A \approx 1.8856\textsf{ sq. units} \end{array}}} \\\\\hline\end{array}$

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5 0

2 years ago

9. If y = 30 when x = 7, find x when y = 12.

Marianna [84]

I have my work at the bottom but to explain you have to divide 30/12, and I got 2.5. That means 30 is 2.5 times bigger than 12. Because we have to find x, we have to divide 7/2.5, to find what is 2.5 times smaller than 7. That’s 2.8. So, x= 2.8

6 0

2 years ago