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3 years ago

15

How do you write 100 as a power of ten

1 answer:

stealth61 [152]3 years ago

3 0

The answer is to that is 10² because 10small one equals 10 and 10³ equals 1000 and you going on my multiplying it by 1ofoe example 100*10 equals 1000.

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Round 345 to one significant number

lisov135 [29]

I think the answer is 300 because 4 is less tvhan 5 and when you round numbers, if the number is less than you leave it or if it is 5 or more you add one more.

8 0

3 years ago

Insert 2 sets of parentheses to make each sentence true: 2 x 14 – 9 – 17 – 14 = 7 (2 x 14) – 9 – (17 – 14) = 7 2 x (14 – 9) + (–

AlexFokin [52]

Answer:

2 × (14 – 9) – (17 – 14) = 7

Step-by-step explanation:

Evaluate the choices to see which is true.

(2 x 14) – 9 – (17 – 14) = 7 ⇒ 28 -9 -3 ≠ 7

2 x (14 – 9) + (– 17 – 14) = 7 ⇒ 2(5) +(-31) ≠ 7

(2 x 14) – (9 – 17) – 14 = 7 ⇒ 28 -(-8) -14 ≠ 7

2 x (14 – 9) – (17 – 14) = 7 ⇒ 2(5)- 3 = 7 . . . . true

3 0

3 years ago

Alice wanted to solve for x by getting rid if the +8. explain and correct Alice error

ra1l [238]

<u>Answer:</u> The value of x will be 2.

<u>Step-by-step explanation:</u>

To calculate the value of 'x' or solve for x, we need to isolate x from 8.

In the equation, 8 is in addition with 'x' so for isolating it, we need to apply the reverse operation of addition.

Reverse operation of addition is subtraction.

We need to subtract 8 from both sides of the equation, we get:

$x+8-8=10-8\\\\x=2$

By doing this, we get the value of 'x' as 2.

Error done by Alice was that she added 8 and it did not led to the isolation of 'x'.

4 0

3 years ago

Read 2 more answers

Why is 6 over 8 greater than 5 over 8 but less than 7 over 8

azamat

Because 6 over 8 is between 7 over 8 and 5 over 8

6 0

3 years ago

Read 2 more answers

Find an equation of the sphere with center (4, −12, 8) and radius 10.

Lorico [155]

Answer:

Step-by-step explanation:

To find an equation of the sphere with center (4, −12, 8) and radius 10

$(x-4)^2+(y+12)^2+(z-8)^2 = 100$

intersection with xy-plane

Put z=0

$(x-4)^2+(y+12)^2+(0-8)^2 = 100$

$(x-4)^2+(y+12)^2 = 36$

(A circle with centre at (4,-12) and radius 6)

intersection with xz-plane

Put y =0

$(x-4)^2+(0+12)^2+(z-8)^2 = 100$

$(x-4)^2+(z-8)^2 = -44$

Sum of squares cannot be positive, so DNE

intersection with yz-plane

Put x=0

$(0-4)^2+(y+12)^2+(z-8)^2 = 100$

$(y+12)^2+(z-8)^2 = 84$

A circle in YZ plane with centre at y =-12 and z =8 and radius square root of 84

5 0

3 years ago