Answer:
43
Step-by-step explanation:
g(x)=4x+3
plug in 10
g(10)= 4(10)+3
= 40+3
= 43
Answer:
2.2 metres squared
Step-by-step explanation:
We need to find the area of this trapezoid.
The area of a trapezoid is denoted by:
, where
and
are the parallel bases and h is the height
Here, we already know the lengths of the two bases; they are 0.9 metres and 2.3 metres. However, we need to find the length of the height.
Notice that one of the angles is marked 45 degrees. Let's draw a perpendicular line from top endpoint of the segment labelled 0.9 to the side labelled 2.3. We now have a 45-45-90 triangle with hypotenuse 2.0 metres. As one of such a triangle's properties, we can divide 2.0 by √2 to get the length of both legs:
2.0 ÷ √2 = √2 ≈ 1.414 ≈ 1.4
Thus, the height is h = 1.4 metres. Now plug all these values we know into the equation to find the area:


The answer is thus 2.2 metres squared.
<em>~ an aesthetics lover</em>
The amount of money Ben had to begin with after spending 1/6 and 1/2 of it is 57 dollars.
<h3>How to find the how much money he had with an equation?</h3>
let
x = amount he had to begin with
He spent 1/6 of his money on a burger, fries, and a drink. Therefore,
amount spent on burger, fries, and a drink = 1 / 6 x
Hence,
amount he had left = x - 1 / 6 x =6x - x /6 = 5 / 6 x
Then he spent half of the money he had left.
1 / 2(5 /6 x) = 5 + 8.25 + 10.50
5 / 12 x = 23.75
cross multiply
5x = 23.75 × 12
5x = 285
divide both sides by 5
x = 285 / 5
x = 57
Therefore, the amount of money he have to begin with is $57.
learn more on equation here: brainly.com/question/5718696
Therefore, cube root of 1000 is 10.
Step-by-step explanation:
= 3√1000
= 3√(10 × 10 × 10)
= 3√(103)
= 10
Answer:
12
Step-by-step explanation:
I did order of operations, GEMA OR PEMDAS, etc.