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4 years ago

Please help, find the area of the composite shape.

Mathematics

1 answer:

vitfil [10]4 years ago

4 0

Answer:

there is no composite shape at the moment...

Step-by-step explanation:

where is the shape?

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Determine the equation

sergejj [24]

Answer:

$y=\frac{3}{4} x-\frac{5}{4}$

Step-by-step explanation:

two points on the line are

(-1, -2)

(-5, -5)

The slope:

$m=\frac{-2-(-5)}{-1-(-5)} =\frac{-2+5}{-1+5} =\frac{3}{4}$

The equation:

with (-1,-2)

$y-(-2)=\frac{3}{4} (x-(-1))$

$y+2=\frac{3}{4} x+\frac{3}{4}$

$y=\frac{3}{4} x+\frac{3}{4} -\frac{8}{4}$

$y=\frac{3}{4} x-\frac{5}{4}$

Hope this helps

8 0

3 years ago

For science class, a student recorded the high and low temperatures, in Fahrenheit, over a ten day period in September. The data

statuscvo [17]

Answer:

hdghjdguwa

Step-by-step explanation:

dznsdgb

5 0

3 years ago

In a box of 10 calculators, one is defective. In how many ways can four calculators be selected, if you know one

Juliette [100K]

Given:

Total number of calculators in a box = 10

Defective calculators in the box = 1

To find:

The number of ways in which four calculators be selected and one of the four calculator is defective.

Solution:

We have,

Total calculators = 10

Defective calculators = 1

Then, Non-defective calculator = 10-1 = 9

Out of 4 selected calculators 1 should be defective. So, 3 calculators are selected from 9 non-defective calculators and 1 is selected from the defective calculator.

$\text{Total ways}=^9C_3\times ^1C_1$

$\text{Total ways}=\dfrac{9!}{3!(9-3)!}\times 1$

$\text{Total ways}=\dfrac{9\times 8\times 7\times 6!}{3\times 2\times 1\times 6!}$

$\text{Total ways}=\dfrac{9\times 8\times 7}{6}$

$\text{Total ways}=3\times 4\times 7$

$\text{Total ways}=84$

Therefore, the four calculators can be selected in 84 ways.

7 0

3 years ago

Helppppppp pleaseeee

Natasha2012 [34]

Answer:

23 more cars than trucks.... It means you add 23 over the number of trucks

2 trucks | 2+23 =25

Do u get it

8 0

3 years ago

I need help PLEASE!!!

Hitman42 [59]

Answer:

1. ∠A and ∠B are right angles. Given

2. m∠A = m∠ B All right angles are congruent.

3. ∠BEC≅ ∠AED Vertical angles are congruent

4. ΔCBE ~ ΔDAE AA

Step-by-step explanation:

A proof always begins with the givens.

1. ∠A and ∠B are right angles. -------------->Given

2. m∠A = m∠ B are equal since-----------> All right angles are congruent.

3. ∠BEC≅ ∠AED are also equal since---->Vertical angles are congruent

4. ΔCBE ~ ΔDAE since two angles are equal----------> AA

7 0

3 years ago