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butalik [34]
3 years ago
11

Hadi earned $358.00 from his job this week. Next week, he expects his pay to be 17% less than it was this week. How much money s

hould Hadi expect to be paid next week?
Mathematics
1 answer:
Novay_Z [31]3 years ago
5 0

17%=0,17

1) 358000×0,17=60860

2) 358000-60860=297140

Answer: He shoud expect $297140 next week

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Use the parabola tool to graph the quadratic function f(x) =-5x2 -2
enot [183]
Your first point would be on (0, 2), the vertex. Then, put a point on (-2, 22). Hope that helps!
3 0
3 years ago
A ditch contains 10 centimeters of water. Rainwater accumulates in the ditch as follows: 10 centimeters of water by the end of t
kondaur [170]
Firstly, let's create a function of f(t) where t represents the time that has past, and f(t) represents the amount of rainwater. We know that when t=1, then f(t)=10, and t=2 then f(t)=15. So, let's take that and analyze it:

(1,10)
(2,15)
m = (15-10)/(2-1) = 5
y-intercept = 5
∴ f(t) = 5t+5

Now we just evaluate t for 10:

f(10) = (5*10)+5
f(10) = 55
3 0
3 years ago
PLZ HELP QUICK PLZ
olya-2409 [2.1K]

Answer:

hope this helps you

5 0
3 years ago
The slope, m, of a linear equation can be found using the formula <img src="https://tex.z-dn.net/?f=m%20%3D%20%5Cfrac%7By_2%20-%
Svetradugi [14.3K]

Answer:

Answer c: m(x2 - x1) + y1 = y2

Step-by-step explanation:

Hi there!

Let´s start writting the formula

m = \frac{y2 -y1}{x2-x1}

If we multiplicate both sides by (x₂ - x₁):

m(x2 - x1) = y2 - y1

Then if we add y₁ to both sides of the equation:

m(x2 - x1) + y1 = y2

Then, the right answer is the answer c)

Have a nice day!

4 0
3 years ago
Answer the question in the picture
nekit [7.7K]

Recall the angle sum identities:

\sin(x+y)=\sin x\cos y+\cos x\sin y

\cos(x+y)=\cos x\cos y-\sin x\sin y

Now,

\tan(x+y)=\dfrac{\sin(x+y)}{\cos(x+y)}=\dfrac{\sin x\cos y+\cos x\sin y}{\cos x\cos y-\sin x\sin y}

Divide through numerator and denominator by \cos x\cos y to get

\tan(x+y)=\dfrac{\tan x+\tan y}{1-\tan x\tan y}

Next, we use the fact that x,y lie in the first quadrant to determine that

\sin x=\dfrac12\implies\cos x=\sqrt{1-\sin^2x}=\dfrac{\sqrt3}2

\cos y=\dfrac{\sqrt2}2\implies\sin x=\sqrt{1-\cos^2x}=\dfrac1{\sqrt2}

So we then have

\tan x=\dfrac{\sin x}{\cos x}=\dfrac{\frac12}{\frac{\sqrt3}2}=\dfrac1{\sqrt3}

\tan y=\dfrac{\sin y}{\cos y}=\dfrac{\frac1{\sqrt2}}{\frac{\sqrt2}2}=1

Finally,

\tan(x+y)=\dfrac{\frac1{\sqrt3}+1}{1-\frac1{\sqrt3}}=\dfrac{1+\sqrt3}{\sqrt3-1}=2+\sqrt3\approx3.73

4 0
3 years ago
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