Find the perimeter of a rectangle is 56 cm. the length of the rectangle is 2 cm more than the width

2 answers:

6 0

P = 56 = 2l + 2w and l=2+w

56 = 2(2+w)+2w
56 = 4 + 2w + 2w
52 = 4w
w = 13 cm
l = 13+2
l = 15 cm

vodka [1.7K]2 years ago

6 0

We call : → l = lenght
We know the perimeter of the
w = width the rectangle : P = 2(w+l)

P = perimeter

P = 2(w+l) but l = w+2

56 = 2(w+w+2)

56 = 2(2w+2)

56 = 4w + 4

4w = 56 - 4

4w = 52 l = w + 2

w = 52 : 4 l = 13 + 2

w = 13 cm l = 15 cm

You might be interested in

What is the degree of the monomial 3x²y3?<br> O2<br> 03<br> 05<br> 06

Natalija [7]

Answer:

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

Find the greatest common factor of x^2y and xy^2

Dimas [21]

1. Find the H.C.F. of 4x2y3 and 6xy2z.
Solution:
The H.C.F. of numerical coefficients = The H.C.F. of 4 and 6.
Since, 4 = 2 × 2 = 22 and 6 = 2 × 3 = 21 × 31
Therefore, the H.C.F. of 4 and 6 is 2

6 0

3 years ago

Tristen is packing lunch boxes for the school trip. Each lunch box consists of 1 sandwich, 1 fruit, and 1 drink. Tristen can cho

Lerok [7]

Step-by-step explanation:

3*4*2=24

Even if he doesn't use them all, he has 24 choices.

These kind of problems are permutation type problem. Order is not critical. Imagine yourself preparing such a meal. You have choices to be made in Sandwiches, fruits and drinks. It does not matter which order you set up the lunch box. It only matters that you chose.

You can't leave one out, nor does the problem allow you to trade (say) 2 drinks for one fruit.

6 0

2 years ago

The weight of laboratory grasshoppers follows a Normal distribution, with a mean of 90 grams and a standard deviation of 2 grams

JulsSmile [24]

Answer:

95.44% of the grasshoppers weigh between 86 grams and 94 grams.

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Mean of 90 grams and a standard deviation of 2 grams.

This means that $\mu = 90, \sigma = 2$