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kodGreya [7K]
3 years ago
12

Simplify the following expressions below using the distributive property and by combining like terms.

Mathematics
1 answer:
Goryan [66]3 years ago
5 0

Answer:

your answers would be : -2+2x+3x-4+3

-3y-2+3y-4

2+3+2y+3x

Step-by-step explanation:

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Please answer the following with Explaination
Zolol [24]

Step-by-step explanation:

brainliest plzz its confirm

5 0
3 years ago
Josh can long-jump 195% of his height with a running start. If Josh is 66 inches tall, how far can he long-jump?
Nikolay [14]

Answer: 128.7 inches

Step-by-step explanation:

Josh's height is 66 inches.

When he jumps with a running start, he can jump as far as 195% of his height.

The length he can jump is therefore 195% multiplied by his height:

= Josh height * 195%

= 66 * 195/100

= 128.7 inches

8 0
3 years ago
The sum of squares of three consecutive numbers is 77 find the numbers​
ollegr [7]

Answer:

4,5,6 are the three consecutive numbers. 16, 25 and 36 are their squares.

Step-by-step explanation:

Let the three consecutive numbers be x, (x+1), (x+2)

Now, the squares of these three numbers are  x^{2} ,(x+1)^{2} ,(x+2)^{2}

Sum = 77

∴by the problem ,

x^{2} +(x+1)^{2}+(x+2)^{2}  = 77\\x^{2} +(x^{2} +2x+1)+(x^{2}+4x +4) = 77\\x^{2} +x^{2} +2x+1+x^{2} +4x +4 = 77\\3x^{2} +6x+5 = 77\\3x^{2} +6x = 77-5\\3x^{2} + 6x = 72\\3x^{2} +6x-72= 0\\

{Taking 3 common }

x^{2} +2x- 24 = 0\\

{By factorization}

x^{2} +2x- 24 =0\\ x^{2} +6x-4x-24=0\\x(x+6)-4(x+6)=0\\(x+6)(x-4)=0\\

Therfore,

x= -6,4\\

<em>X can't be negetive </em>

∴ x =4\\x+1=5\\x+2=6

The squares of the three consecutive numbers are 16, 25, 36

The three consecutive numbers whose sum is 77 are 4, 5, 6

3 0
4 years ago
4) Write down all the subsets of (a) {p, q}​
antiseptic1488 [7]

Answer:  

If P is a subset of Q, then by definition, every element of P is an element of Q. So, as Q contains P, every element in either P or Q is in Q, so the union is Q. For the second, every element of P and Q is in P, so the intersection is P.

Step-by-step explanation:

5 0
3 years ago
Inequality to x2 &lt; 64
Anarel [89]
              NOT MY WORDS TAKEN FROM A SOURCE!

(x^2) <64  => (x^2) -64 < 64-64 => (x^2) - 64 < 0 64= 8^2    so    (x^2) - (8^2) < 0  To solve the inequality we first find the roots (values of x that make (x^2) - (8^2) = 0 ) Note that if we can express (x^2) - (y^2) as (x-y)* (x+y)  You can work backwards and verify this is true. so let's set (x^2) - (8^2)  equal to zero to find the roots: (x^2) - (8^2) = 0   => (x-8)*(x+8) = 0       if x-8 = 0 => x=8      and if x+8 = 0 => x=-8 So x= +/-8 are the roots of x^2) - (8^2)Now you need to pick any x values less than -8 (the smaller root) , one x value between -8 and +8 (the two roots), and one x value greater than 8 (the greater root) and see if the sign is positive or negative. 1) Let's pick -10 (which is smaller than -8). If x=-10, then (x^2) - (8^2) = 100-64 = 36>0  so it is positive
2) Let's pick 0 (which is greater than -8, larger than 8). If x=0, then (x^2) - (8^2) = 0-64 = -64 <0  so it is negative3) Let's pick +10 (which is greater than 10). If x=-10, then (x^2) - (8^2) = 100-64 = 36>0  so it is positive Since we are interested in (x^2) - 64 < 0, then x should be between -8 and positive 8. So  -8<x<8 Note: If you choose any number outside this range for x, and square it it will be greater than 64 and so it is not valid.

Hope this helped!

:)
7 0
3 years ago
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