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Zigmanuir [339]
3 years ago
10

Solve -5 √x =-15 A. X= -9 B. X= -7 C. X= 9 D. X= 7

Mathematics
2 answers:
Leokris [45]3 years ago
8 0
Divide both sides by -5
You would then get (square root)x=3 since the negatives would cancel out
Then you square both sides
You will be left with (x)*2=(3)*2
Your answer will be C. X=9
Hope that helped!
avanturin [10]3 years ago
6 0
X = 9

The answer is C
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If g(x) is a linear function such that g(-3) = 2 and g(1) = -4, find g(7).
eimsori [14]
<h3>Answer:   -13</h3>

=======================================

Explanation:

g(-3) = 2 means x = -3 and y = 2 pair up together to form the point (-3,2)

g(1) = -4 means we have the point (1,-4)

Find the slope of the line through the two points (-3,2) and (1,-4)

m = (y2-y1)/(x2-x1)

m = (-4-2)/(1-(-3))

m = (-4-2)/(1+3)

m = -6/4

m = -3/2

m = -1.5

The general slope intercept form y = mx+b turns into y = -1.5x+b after replacing m with -1.5

Plug in (x,y) = (-3,2) which is one of the points mentioned earlier and we end up with this new equation:  2 = -1.5*(-3) + b

Let's solve for b

2 = -1.5*(-3)+b

2 = 4.5 + b

2-4.5 = 4.5+b-4.5 .... subtract 4.5 from both sides

-2.5 = b

b = -2.5

Therefore, y = mx+b becomes y = -1.5x-2.5 meaning the g(x) function is g(x) = -1.5x-2.5

The last step is to plug in x = 7 and compute

g(x) = -1.5*x - 2.5

g(7) = -1.5*7 - 2.5

g(7) = -10.5 - 2.5

g(7) = -13

6 0
2 years ago
Read 2 more answers
The cosine ratio is used to find missing angle measures and side measures in
VashaNatasha [74]
... right triangles.
7 0
3 years ago
the half-life of chromium-51 is 38 days. If the sample contained 510 grams. How much would remain after 1 year?​
madam [21]

Answer:

About 0.6548 grams will be remaining.  

Step-by-step explanation:

We can write an exponential function to model the situation. The standard exponential function is:

f(t)=a(r)^t

The original sample contained 510 grams. So, a = 510.

Each half-life, the amount decreases by half. So, r = 1/2.

For t, since one half-life occurs every 38 days, we can substitute t/38 for t, where t is the time in days.

Therefore, our function is:

\displaystyle f(t)=510\Big(\frac{1}{2}\Big)^{t/38}

One year has 365 days.

Therefore, the amount remaining after one year will be:

\displaystyle f(365)=510\Big(\frac{1}{2}\Big)^{365/38}\approx0.6548

About 0.6548 grams will be remaining.  

Alternatively, we can use the standard exponential growth/decay function modeled by:

f(t)=Ce^{kt}

The starting sample is 510. So, C = 510.

After one half-life (38 days), the remaining amount will be 255. Therefore:

255=510e^{38k}

Solving for k:

\displaystyle \frac{1}{2}=e^{38k}\Rightarrow k=\frac{1}{38}\ln\Big(\frac{1}{2}\Big)

Thus, our function is:

f(t)=510e^{t\ln(.5)/38}

Then after one year or 365 days, the amount remaining will be about:

f(365)=510e^{365\ln(.5)/38}\approx 0.6548

5 0
2 years ago
Solve triangle ABC if AB=24.6, BC=29.7, and CA=30
PtichkaEL [24]
Triangle ABC equals 84.3
8 0
3 years ago
Mrs. Jones decided to buy some pencils for her class. She bought 5 packages of pencils, and each package contained 42 pencils. T
lisov135 [29]

Answer:42x5=210/14=15

Step-by-step explanation:Each student would get 15 pencils

3 0
2 years ago
Read 2 more answers
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