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4 years ago

7

The width of a rectangle is the length minus 2 units. The area of the rectangle is 35 units. What is the width, in units, of the

rectangle?

1 answer:

Andrews [41]4 years ago

6 0

Answer: width = 5 units

Step-by-step explanation:

Let L represent the length of the rectangle.

The width of a rectangle is the length minus 2 units. It means that the width of the rectangle is (L - 2) units.

The formula for determining the area of a rectangle is expressed as

Area = length × width

The area of the rectangle is 35 units. It means that

L(L - 2) = 35

L² - 2L = 35

L² - 2L - 35 = 0

L² + 5L - 7L - 35 = 0

L(L + 5) - 7(L + 5) = 0

L - 7 = 0 or L + 5 = 0

L = 7 or L = - 5

Since the length cannot be negative, then

L = 7 units

Width = 7 - 2 = 5 units

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Ms. Van Tress has 134 students. She sees 38 students before lunch and 96 students after lunch.

liubo4ka [24]

Answer:

<em /> $\%Percent = 28\%$ <em> ---- Before Lunch</em>

<em /> $\%Percent = 72\%$ <em> --- After Lunch</em>

<em />

Step-by-step explanation:

Given

$Total = 134$

$Before\ Lunch = 38$

$After\ Lunch = 96$

Required

Determine the percentage of students seen in both cases.

This percentage is calculated by dividing the number of students seen by the total number of students. Then multiplied by 100%

i.e

$\%Percent = \frac{Seen}{Total} * 100\%$

- Before Lunch;

$\%Percent = \frac{Before\ Lunch}{Total} * 100\%$

$\%Percent = \frac{38}{134} * 100\%$

$\%Percent = \frac{3800\%}{134}$

<em /> $\%Percent = 28\%$ <em> --- approximated</em>

- After Lunch;

$\%Percent = \frac{After\ Lunch}{Total} * 100\%$

$\%Percent = \frac{96}{134} * 100\%$

$\%Percent = \frac{9600\%}{134}$

$\%Percent = 72\%$ <em>--- approximated</em>

3 0

3 years ago

The answer is 40, I hope this helps!

Maurinko [17]

Answer:

thank you

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

A 5-pound bag of apples costs $3.00. At the

NeX [460]

Answer:

$4.20

Step-by-step explanation:

Setup a ratio.

5/3=7/x

5x=7(3)

x=21/5

x=4.2 or for this problem 4.20

7 0

4 years ago

The weights of certain machine components are normally distributed with a mean of 8.04 g and a standard deviation of 0.08 g. Fin

NISA [10]

Answer:

The bottom 3 is separated by weight 7.8896 g and the top 3 is separated by weight 8.1904 g.

Step-by-step explanation:

We are given that

Mean, $\mu=8.04 g$

Standard deviation, $\sigma=0.08g$

We have to find the two weights that separate the top 3% and the bottom 3%.

Let x be the weight of machine components

$P(Xx_2)=0.03$

$P(X$

=0.03

From z- table we get

$P(Z1.88)=0.03$

Therefore, we get

$\frac{x_1-8.04}{0.08}=-1.88$

$x_1-8.04=-1.88\times 0.08$

$x_1=-1.88\times 0.08+8.04$

$x_1=7.8896$

$\frac{x_2-8.04}{0.08}=1.88$

$x_2=1.88\times 0.08+8.04$

$x_2=8.1904$

Hence, the bottom 3 is separated by weight 7.8896 g and the top 3 is separated by weight 8.1904 g.

5 0

3 years ago

A un paciente se le toma la temperatura a las 7 am, 9 am, 2 pm y a las 4 pm. La primera segunda y cuarta lecturas son de 66 grad

Gnesinka [82]

Answer:

94

Step-by-step explanation:

(Lo siento, español no es mi primero lingua, pero intentaré mi mejor :) )

La ecuación is (66 + 90 + x + 94)/4 = 86

Multiple los dos de los lado por 4.

66 + 90 + x + 94 = 344

Hace 66 + 90 + 94, y ponelos a la otro lado con sustracción.

X = 94

94 es la tercera lecture.

7 0

3 years ago