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inessss [21]
4 years ago
12

Suppose x=10 and y=10. what is x after evaluating the expression (y>= 10)||(x++ >10)? java

Mathematics
1 answer:
Mamont248 [21]4 years ago
4 0
<span>Suppose x=10 and y=10. what is x after evaluating the expression (y>= 10)||(x++ >10)? java
Answer is 10</span>
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fenix001 [56]

Answer:

10=(5x0.75)+6

Step-by-step explanation:

you'll still have 25¢ however

4 0
3 years ago
If a rectangle has a length of 12 inches and a width of inches, what is the value of its perimeter ? Include units.
kiruha [24]

Answer:

36

Step-by-step explanation:

3 0
3 years ago
You work 35 hours/week for 52 weeks and are given the option to be paid hourly or to go on salary. In which situation will you e
rjkz [21]

Answer:

Hourly

Step-by-step explanation:

if you work for that long and for that many hours you would get alot of money in about a month or two.

4 0
4 years ago
Read 2 more answers
The scatter plot shows the number of customers in a restaurant for hours of the dinner services on two different Saturday nights
charle [14.2K]

A. The value y-intercept represent in the context of the problem is 50.

B. The slope of the line is -10.

C. The equation to represent the line is y= -10x+50.

D. The number of customers in the restaurant at 8:30pm is 35.

<u>Step-by-step explanation:</u>

The equation of the line is given by y=mx+c.

y is the number of customers and x is the hours.

7.00 p.m. is considered as 0.

To find the slope m =\frac{rise}{run}.

rise = y_{2}-y_{1}.

run = x_{2}-x_{1}.

From the graph, choose two points. Let us take (0,50) and (4,10).

m=\frac{-40}{4}.

m=-10.

∴The equation can be written as y= -10x+c.

To find the y-intercept value c, substitute any point.

Let us substitute(0,50) for instance,

50=0+c.

c=50.

The equation for the given graph is y= -10x+50.

To find the customers at the restaurant at 8.30 pm.

we know that x=0 at 7.00 pm,

∴ x=1.5 can be considered as 8.30 pm.

Substitute x=1.5 in the line equation.

y= -10(1.5)+50.

y= -15-50.

y= 35.

5 0
3 years ago
Prove the following identity :<br>Sin α . Cos α .<br> Tan α =  (1 – Cos<br> α)  (1 + Cos<br> α)
Umnica [9.8K]
Let's work on the left side first. And remember that
the<u> tangent</u> is the same as <u>sin/cos</u>.

sin(a) cos(a) tan(a)

Substitute for the tangent:

[ sin(a) cos(a) ] [ sin(a)/cos(a) ]

Cancel the cos(a) from the top and bottom, and you're left with

[ sin(a) ] . . . . . [ sin(a) ] which is [ <u>sin²(a)</u> ]  That's the <u>left side</u>.

Now, work on the right side:

[ 1 - cos(a) ] [ 1 + cos(a) ]

Multiply that all out, using FOIL:

[ 1 + cos(a) - cos(a) - cos²(a) ]

= [ <u>1 - cos²(a)</u> ] That's the <u>right side</u>.

Do you remember that for any angle, sin²(b) + cos²(b) = 1  ?
Subtract cos²(b) from each side, and you have sin²(b) = 1 - cos²(b) for any angle.

So, on the <u>right side</u>, you could write [ <u>sin²(a)</u> ] .

Now look back about 9 lines, and compare that to the result we got for the <u>left side</u> .

They look quite similar. In fact, they're identical. And so the identity is proven.

Whew !





4 0
3 years ago
Read 2 more answers
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