Answer:
Cost of small box = $13
Cost of large box = $18
Step-by-step explanation:
Let the cost of small box of candy = $s
And the cost of large candy box = $b
Beth sold 9 small and 10 large boxes of candies for $297.
9s + 10b = 297 -----(1)
Sarah sold 8 small and 5 large boxes of candies for $194.
8s + 5b = 194 -------(2)
Multiply equation (2) by 2 then subtract equation (1) from (2).
2(8s + 5b) - (9s + 10b) = (2×194) - 297
16s + 10b - 9s - 10b = 388 - 297
7s = 91 ⇒ s = $13
From equation (2),
(8×13) + 5b = 194
104 + 5b = 194
5b = 90
b = $18
Translation (1,1) as it has moved one left horizontally and one up vertically. that is if the question is about the bottom left shape. if it is about the top right then it would be (-1,-1)
<span>The answer is (x</span>¹⁰<span>y</span>¹⁴<span>)/729.
Explanation:
We can begin simplifying inside the innermost parentheses using the properties of exponents. The power of a power property says when you raise a power to a power, you multiply the exponents. This gives us
[(3</span>³<span>x</span>³<span>y</span>⁻¹⁵<span>)/(x</span>⁸<span>y</span>⁻⁸<span>)]</span>⁻²<span>.
Negative exponents tell us to "flip" sides of the fraction, so within the parentheses we have
[(3</span>³<span>x</span>³<span>y</span>⁸<span>)/(x</span>⁸<span>y</span>¹⁵<span>)]</span>⁻²<span>.
Using the quotient property, we subtract exponents when dividing powers, which gives us
(3</span>³<span>/x</span>⁵<span>y</span>⁷<span>)</span>⁻²<span>.
Evaluating 3</span>³<span>, we have
(27/x</span>⁵<span>y</span>⁷<span>)</span>⁻²<span>.
Using the power of a power property again, we have
27</span>⁻²<span>/x</span>⁻¹⁰<span>y</span>⁻¹⁴<span>.
Flipping the negative exponents again gives us x</span>¹⁰<span>y</span>¹⁴<span>/729.</span>
Answer:
Step-by-step explanation:
It is given that,
V = bdt i.e. V is equal to the product of b, d and t
We need to find the formula for d.
Dividing both sides of the given expression by bt
So,
Hence, is the value of d.
Step-by-step explanation:
Q1 . (f+g)(x) = f(x) + g(x)
=4x-4 +2x^2 -3x
= 2x^2 + x -4
Q2. (f-g)(x) = f(x) - g(x)
= 2x^2−2 - (4x+1)
= 2x^2 -2 -4x -1
= 2x^2 - 4x -3
Q3. h(x)=3x−3 and g(x)=x^2+3
(h.g)(x) = h(x) × g(x)
= (3x-3) × (x^2 + 3)
=3x^3 -3x^2 + 9x -9
Q4.f(x)=x+4 and g(x)=x+6
(f/g)(x) = f(x) ÷ g(x)
= x+4 / x+6
the domain restriction is x>-6
x<-6
x doesn't equal (-6)
