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3 years ago

True or false, there are two solutions to the equation |x-21|= -7

Mathematics

2 answers:

Alex Ar [27]3 years ago

6 0

False, there are no solutions to the equation |x − 21| = −7 because the absolute value cannot be less than 0.

notsponge [240]3 years ago

4 0

Answer:

false

Step-by-step explanation:

You might be interested in

Mean of four numbers a ,b, c, and d is 8 and mean of seven

Nana76 [90]

Answer:

17 1/3 or 17.33

Step-by-step explanation:

Mean of four numbers a, b, c, and d is 8:

(a + b + c + d)/4 = 8
⇒ a + b + c + d = 32

Mean of seven numbers x, y, z, a, b, c and d is 12:

(x + y + z + a + b + c + d)/7 = 12

Substitute the sum of 4 numbers:

x + y + z + 32 = 7*12
x + y + z = 84 - 32
x + y + z = 52

Mean of x, y and z is:

52/3 = 17 1/3 or 17.33

4 0

2 years ago

Choose the correct term that makes the sentence true. The (greatest) (or) (least) of the common factors of two or more numbers i

stiv31 [10]

the greatest is 10 and the least is one but I don't know for sure but I don't know if that what you looking for

5 0

3 years ago

A standard American Eskimo dog has a mean weight of 30 pounds with a standard deviation of 2 pounds. Assuming the weights of sta

nalin [4]

Answer:

Option 2 - Approximately 24–36 pounds

Step-by-step explanation:

Given : A standard American Eskimo dog has a mean weight of 30 pounds with a standard deviation of 2 pounds. Assuming the weights of standard Eskimo dogs are normally distributed.

To find : What range of weights would 99.7% of the dogs have?

Solution :

The range of 99.7% will lie between the mean ± 3 standard deviations.

We have given,

Mean weight of Eskimo dogs is $\mu=30$

Standard deviation of Eskimo dogs is $\sigma=2$

The range of weights would 99.7% of the dogs have,

$R=\mu\pm3\sigma$

$R=30\pm3(2)$

$R=30\pm6$

$R=30+6,30-6$

$R=36,24$

Therefore, The range is approximately, 24 - 36 pounds.

So, Option 2 is correct.

5 0

3 years ago

Pls help I will give brainliest

miss Akunina [59]

Answer:

Step-by-step explanation:

the distance around the figure--we need to find the perimeter

we have rectangle that has the side, 8 and 15

P(r)=2(8)+2(15)=16+30=46

we also have half of the circle with the diameter =8 and r=4

Circumference =2*3.14*r

C=2*3.14*4=25.12

we need just half of the circle 25.12/2=12.56

P figure=46+12.56=58.56

Area= rectangle=l*w=8*15=120

Area(circle)=3.14*r^2=3.14*16=50.24

half circle 50.24/2=25.12

Area figure=120+25.12=145.12

6 0

3 years ago

Based on the information below what are the values of x and y of the solution to the system of equations used to create the info

Grace [21]

Answer:

(D). x = 2 , y = - 5

Step-by-step explanation:

A = - 8 + 48 = 40

$A_{x}$ = -76 + 156 = 80

$A_{y}$ = 104 - 304 = - 200

x = $\frac{A_{x} }{A}$ = 2

y = $\frac{A_{y} }{A}$ = - 5

( 2, - 5 )

3 0

2 years ago

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