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2 years ago

(Gonna try this again because the last one failed and I really need help..)

Mathematics

1 answer:

antoniya [11.8K]2 years ago

4 0

I think its C. hope this helps!

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Use an area model to solve the<br> expression 0.6 × 0.4. (6-4)

Vinvika [58]

Step-by-step explanation:

the answer is 48 because

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3 years ago

Daniel can by a particular brand of juice in a 6 ounce bottle or a 10 ounce bottle the juice in a 10 ounce bottle contains 150 c

olasank [31]

Answer: There are 90 calories in a 6-ounce bottle.

Step-by-step explanation:

We assume that the quantity of juice in the bottle is directly proportional to the number of calories it contains.

Given: 10 ounces of juice has 150 calories.

i.e. 1 ounce of juice has $\dfrac{150}{10}$ calories.

i.e. 1 ounce of juice has 15 calories.

Now, In 6 ounce of juice, the number if calories = 15 x 6 = 90 calories

Hence, there are 90 calories in a 6-ounce bottle.

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3 years ago

If a = -35, b = 10 cm and c = -5, verify that:

Murljashka [212]

Step-by-step explanation:

(i). a+(b+c) = (a+b)+c

-35+(10-5) = (-35+10)+(-5)

-35+5 = -25-5

-30 = -30

(ii). a×(b+c) = a×b + a×c

-35 × [10+(-5)] = -35×10 + -35×-5

-35 × (10-5) = -350 + 175

-35 × 5 = -350 + 175

-175 = -175

7 0

2 years ago

Find an equation for the inverse of the relation y=6x-3. Show how you found the inverse equation.

Elis [28]

Hi there! The answer is y = 1/6x + 1/2

To find the inverse function x and y need to switch places.

$y = 6x - 3 \\ x = 6y - 3$

Now rewrite the equation, until it will be in standard form (slope-intercept form) again.
$x = 6y - 3$

Add 3.
$x + 3 = 6y$

Switch sides.
$6y = x + 3$

Divide both sides by 6.
$y = \frac{1}{6} x+ \frac{3}{6} \\ y = \frac{1}{6} x + \frac{1}{2}$

Hence, the answer is y = 1/6x + 1/2.
~ Hope this helps you!

4 0

3 years ago

Which is the value of this expression when p=-2 and q=-1?

saul85 [17]

Answer:

D. 4

Step-by-step explanation:

$[(p^2) (q^{-3}) ]^{-2}.[(p)^{-3}(q)^5] ^{-2}\\\\=[(p^2) (q^{-3}) \times(p)^{-3}(q)^5 ]^{-2}\\\\=[(p^{2}) \times(p)^{-3} \times (q^{-3}) \times(q)^5 ]^{-2}\\\\=[(p^{2-3}) \times (q^{5-3}) ]^{-2}\\\\=[(p^{-1}) \times (q^{2}) ]^{-2}\\\\=(p^{-1\times (-2)}) \times (q^{2\times (-2) }) \\\\=p^{2}\times q^{-4} \\\\= \frac{p^2}{q^4}\\\\= \frac{(-2)^2}{(-1)^4}\\\\= \frac{4}{1}\\\\= 4$

8 0

3 years ago