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2 years ago

Find area AND perimeter of the shape

Mathematics

2 answers:

den301095 [7]2 years ago

8 0

Step-by-step explanation:

for the perimeter...

18-6=12

that's to get the slant side

perimeter=10+6+12+12+20

=60

for the area I'm not sure how to solve it sorry!!

Rashid [163]2 years ago

6 0

Answer:

10 x 6 = 60

20 + 10 = 30, 30 x 12 x 1/2 = 180

180 + 60 = 240 cm2 is your area

6 + 6 = 12, 12 + 10 = 22, 22 + 20 = 42

Now lets find hypotenuse:

10 - 20 = 10, 10 divided by 2 = 5

5^2 + 12^2

25 + 144 = 169

Square root of 169 is 13

Add perimeters:

42 + 13 = 55 cm is your perimeter.

You might be interested in

Three consecutive integers have a sum of 42. Find the integers.

faltersainse [42]

Answer:

13, 14, 15

Step-by-step explanation:

Three consecutive integers have a sum of 42 means which 3 numbers in a row add up to 42

We can call the lowest integer x,

the next integer x+1 (since our integers need to be even)

the third integer x+2

algebraic equation: 42 = x + (x+1) + (x+2).

Combine like terms on the right side,

42 = 3x + 3

Subtract 3 from each side

39 = 3x

Divide each side by 3

x=13, which is our lowest integer.

The next two integers we find with 13 + 1 =.14 and 13 + 2 = 15

13 + 14 + 15 = 42.

6 0

1 year ago

Factorize: x^2 + (a^2+1)/a x+1

In-s [12.5K]

This is what I get not sure if the problem on numerator is together if it is then . X^2+(a^2+1) x^2+a^2+1. -----------------= ----------------- AX+1. Ax+1 Answer: a^2+x^2+1 ---------------- AX+1 If I wrote the actual problem wrong let me know. This is what I understood.

5 0

3 years ago

Factor 3/4 out of 3/4z + 6

Alenkasestr [34]

ANSWER

$\frac{3}{4} z + 6 = \frac{3}{4} (z + 8 )$

EXPLANATION

We want to factor
$\frac{3}{4}$
out of

$\frac{3}{4} z + 6$

The first term is already having a factor of
$\frac{3}{4}$
The constant term which is the second term is not having a factor of
$\frac{3}{4}$
so we need to use a trick of multiplying and dividing by the same factor to obtain,

$\frac{3}{4} z + 6 = \frac{3}{4} z + \frac{3}{4} \times \frac{6}{ \frac{3}{4} }$

We can now factor
$\frac{3}{4}$
out of the right hand side to obtain,

$\frac{3}{4} z + 6 = \frac{3}{4} (z + \frac{6}{ \frac{3}{4} } )$
Let us rewrite the right most fraction using the normal division symbol.

$\frac{3}{4} z + 6 = \frac{3}{4} (z + 6 \div \frac{3}{4} )$

We simplify to obtain,

$\frac{3}{4} z + 6 = \frac{3}{4} (z + 6 \times \frac{4}{3} )$

This further gives us,

$\frac{3}{4} z + 6 = \frac{3}{4} (z + 2 \times 4 )$

$\frac{3}{4} z + 6 = \frac{3}{4} (z + 8 )$

5 0

3 years ago

Read 2 more answers

3x + (-2 + 4x)<br> X = 5

KATRIN_1 [288]

3(5)+(-2+4(5))
15+(-2+20)
15+(18)
15+18
=33

7 0

3 years ago

A large pool of adults earning their first driver’s license includes 50% low-risk drivers, 30% moderate-risk drivers, and 20% hi

Mamont248 [21]

Answer:

The probability that these four will contain at least two more high-risk drivers than low-risk drivers is 0.0488.

Step-by-step explanation:

Denote the different kinds of drivers as follows:

L = low-risk drivers

M = moderate-risk drivers

H = high-risk drivers

The information provided is:

P (L) = 0.50

P (M) = 0.30

P (H) = 0.20

Now, it given that the insurance company writes four new policies for adults earning their first driver’s license.

The combination of 4 new drivers that satisfy the condition that there are at least two more high-risk drivers than low-risk drivers is:

S = {HHHH, HHHL, HHHM, HHMM}

Compute the probability of the combination {HHHH} as follows:

P (HHHH) = [P (H)]⁴

= [0.20]⁴

= 0.0016

Compute the probability of the combination {HHHL} as follows:

P (HHHL) = ${4\choose 1}$ × [P (H)]³ × P (L)

= 4 × (0.20)³ × 0.50

= 0.016

Compute the probability of the combination {HHHM} as follows:

P (HHHL) = ${4\choose 1}$ × [P (H)]³ × P (M)

= 4 × (0.20)³ × 0.30

= 0.0096

Compute the probability of the combination {HHMM} as follows:

P (HHMM) = ${4\choose 2}$ × [P (H)]² × [P (M)]²

= 6 × (0.20)² × (0.30)²

= 0.0216

Then the probability that these four will contain at least two more high-risk drivers than low-risk drivers is:

P (at least two more H than L) = P (HHHH) + P (HHHL) + P (HHHM)

+ P (HHMM)

= 0.0016 + 0.016 + 0.0096 + 0.0216

= 0.0488

Thus, the probability that these four will contain at least two more high-risk drivers than low-risk drivers is 0.0488.

6 0

3 years ago