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Bumek [7]
4 years ago
5

Determine wether the given number is a solution of the given equation 10;4-8x=-69

Mathematics
2 answers:
I am Lyosha [343]4 years ago
5 0
No because if you subtract 4 from -69 it equals -73. then if you put -73 divided by -8 it equals positive 9.125.

Or if you put in 10 where the x is you would do 4-8 *10=-64
                                                                                 4-80=-69
                                                                                   -76=-69
otez555 [7]4 years ago
4 0
4 - 8x = - 69
-8x = -69 - 4
-8x = - 73
x = -73/-8
x = 9.125.....no, 10 is not a solution

or, instead of solving for x, we can just sub in 10 for x and see if it makes the equation true.

4 - 8x = -69
4 - 8(10) = -69
4 - 80 = -69
-76 = -69 (incorrect...therefore, it is not a solution)
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The total area of Canada is 9 984 670 km2. Calculate the side length of a square with the same area. Show your work.
Korolek [52]

Answer:

3159.853 km

Step-by-step explanation:

The question says a square has an area of 9,984,670 km^2, which means that all sides are the same length. So you can use the equation:

9984670 = (x)(x) which rearranges to 998470 = x^2

get the square root and you get:

x = 3159.853 km

8 0
3 years ago
If p(x) = x2 – 1 and q(x) = 5(x-1) which expression is equivalent to (p – q)(x)?
lilavasa [31]

Answer:

Option C is correct.

The expression which is equivalent to (p-q)(x)  is, x^2 -1 - 5(x-1)

Step-by-step explanation:

Given: p(x) = x^2 -1 and q(x) = 5(x-1)

To find the expression which is equivalent to (p-q)(x):

here p and q both are function assuming;

we can write the expression :

(p-q)(x) = p(x) - q(x) = x^2 -1 - 5(x-1)

therefore, the expression which is equivalent to (p-q)(x)  is;   x^2 -1 - 5(x-1)

3 0
4 years ago
Read 2 more answers
rock is thrown upward with a velocity of 27 meters per second from the top of a 23 meter high cliff and it misses the cliff on t
ahrayia [7]

the rock will be at 11 meters from the ground level after 5.92 seconds

Step-by-step explanation:

The motion of the rock is a free-fall motion, since the rock is acted upon the force of gravity only. Therefore, it is a uniformly accelerated motion, so its position at time t is given by the equation:

y=h+ut+\frac{1}{2}at^2

where

h = 23 m is the initial height

u = 27 m/s is the initial velocity, upward

a=g=-9.8 m/s^2 is the acceleration of gravity, downward

t is the time

We want to find the time t at which the position of the rock is

y = 11 m

Substituting and re-arranging the equation, we find

11=23+27t-4.9t^2\\4.9t^2-27t-12=0

This is a second-order equation, which has solutions:

t=\frac{27\pm \sqrt{(-27)^2-4(4.9)(-12)}}{2(4.9)}=\frac{27\pm \sqrt{964.2}}{9.8}

So

t_1 = -0.41 s

t_2=5.92 s

The first solution is negative so we neglect it: therefore, the rock will be at 11 meters from the ground level after 5.92 seconds.

Learn more about free fall:

brainly.com/question/1748290

brainly.com/question/11042118

brainly.com/question/2455974

brainly.com/question/2607086

#LearnwithBrainly

8 0
3 years ago
Are the ratios 1:2 and 2:3 equivalent?
timofeeve [1]

Answer:

Step-by-step explanation:

No

8 0
3 years ago
Read 2 more answers
Please help me ASAP!Please this is due soon and i don't have all day.
Helga [31]

<u><em>Answer:</em></u>

Section 1: 5.59%

Section 2: -2.16%

Section 3: -10.82%

Section 4: -6.54%

Section 5: -2.16%

Section 6: 7.86%

<u><em>Explanation:</em></u>

<u>The formula to calculate the percent change in mass is given as:</u>

PercentChange = \frac{ChangeInMass}{StartingMass}*100=\frac{NewMass - OldMass}{OldMass}*100

<u>1- Section 1:</u>

Old mass = 7.83 g and New mass = 8.268 g

PercentChange = \frac{8.268-7.83}{7.83}*100 =5.59%

<u>2- Section 2:</u>

Old mass = 7.4 g and New mass = 7.24 g

PercentChange = \frac{7.24-7.4}{7.4}*100 =-2.16%

<u>3- Section 3:</u>

Old mass = 7.3 g and New mass = 6.51 g

PercentChange = \frac{6.51-7.3}{7.3}*100 =-10.82%

<u>4- Section 4:</u>

Old mass = 7.49 g and New mass = 7.0 g

PercentChange = \frac{7.0-7.49}{7.49}*100 =-6.54%

<u>5- Section 5:</u>

Old mass = 7.4 g and New mass = 7.24 g

PercentChange = \frac{7.24-7.4}{7.7.4}*100 =-2.16%

<u>6- Section 6:</u>

Old mass = 7.89 g and New mass = 8.51 g

PercentChange = \frac{8.51-7.89}{7.89}*100 =7.86%

Hope this helps :)

6 0
3 years ago
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