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mel-nik [20]
3 years ago
7

Clara solved the equation 7 3 x = − 2 3 as shown. 7 3 x ( 3 7 ) = − 2 3 ( 3 7 ) x = −14 What is Clara’s error? Clara should have

divided by StartFraction 3 over 7 EndFraction. Clara should have added StartFraction 7 over 3 EndFraction. Clara should have multiplied by StartFraction 7 over 3 EndFraction. The solution is Negative StartFraction 2 over 7 EndFraction, not –14.
Mathematics
2 answers:
Sphinxa [80]3 years ago
5 0
Answer:15 hope this helps
hjlf3 years ago
4 0

Answer:

15

Step-by-step explanation:

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Is it 12 chairs per three tables? I might be wrong..

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A projectile is fired into the air with an initial vertical velocity of 160 ft/sec from ground level. How many seconds later doe
djverab [1.8K]

The maximum height of the projectile is the maximum point that can be gotten from the projectile equation

The projectile reaches the maximum height after 5 seconds

The function is given as:

\mathbf{h(t) = -16t^2 + 160t}

Differentiate the function with respect to t

\mathbf{h'(t) = -32t + 160}

Set to 0

\mathbf{h'(t) = -32t + 160 = 0}

So, we have:

\mathbf{-32t + 160 = 0}

Collect like terms

\mathbf{-32t =- 160 + 0}

\mathbf{-32t =- 160}

Solve for t

\mathbf{t = \frac{- 160}{-32}}

\mathbf{t = 5}

Hence, the projectile reaches the maximum after 5 seconds

Read more about maximum values at:

brainly.com/question/6636648

8 0
3 years ago
What are the solutions of x^2+10x+21=0?
Luden [163]
(x+7)(x+3) so therefore you set each equation = 0...
x + 7 = 0
x + 3= 0
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6 0
3 years ago
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Do the sides 9, 12 and 15 make a right triangle?
Fynjy0 [20]

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YES

Step-by-step explanation:

because they are all different lengths  

6 0
3 years ago
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In 3-4 sentences, describe how you would find a line parallel to the line 2x + 5y = 15 that goes through the point (-10, 1). Be
larisa86 [58]

the parallel line is 2x+5y+15=0.

Step-by-step explanation:

ok I hope it will work

soo,

Solution

given,

given parallel line 2x+5y=15

which goes through the point (-10,1)

now,

let 2x+5y=15 be equation no.1

then the line which is parallel to the equation 1st

2 x+5y+k = 0 let it be equation no.2

now the equation no.2 passes through the point (-10,1)

or, 2x+5y+k =0

or, 2*-10+5*1+k= 0

or, -20+5+k= 0

or, -15+k= 0

or, k= 15

putting the value of k in equation no.2 we get,

or, 2x+5y+k=0

or, 2x+5y+15=0

which is a required line.

6 0
1 year ago
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