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2 years ago

7

From the beginning of the day the temperature rose 15° so that by 4 pm it was 87°. what was the temperature at the start of the

day?

1 answer:

katrin [286]2 years ago

3 0

The temp started at 72 degrees

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Josh And ash order 2 whole pizzas to eat.Josh ate 5/8 slices of pizza and ash ate 6/8 slices of pizza.Who much pizza is still le

Elenna [48]

2-5/8-6/8=16/8-5/8-6/8=5/8 pizza slices left

3/8+7/8=10/8=1 2/8 pizza slices altogether

hope it helps:))

8 0

3 years ago

You are looking at two area rugs for your bedroom. One is a square with a side length of 6.3 feet. The other is a rectangle that

Dmitry [639]

-----------------------------------------------------------------
Square Rug
-----------------------------------------------------------------
Area = Length x Length
Area = 6.3 x 6.3
Area = 39.69 ft²

-----------------------------------------------------------------
Rectangle Rug
-----------------------------------------------------------------
Area = Length x Width
Area = 4.8 x 8.2
Area = 39.36 ft²

39.69 ft² is bigger than 29.36 ft², so the square rug is bigger.

-----------------------------------------------------------------
Ans: The Square Rug covers a greater area
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8 0

3 years ago

EXPLAIN why we placed the value of x= 4/3( the minimum value) into the equ of gradient(dy/dx) [in the answer, marking scheme att

aliina [53]

$y=x(x-2)^2$
$\implies y'=(x-2)^2+2x(x-2)=3x^2-8x+4=(3x-2)(x-2)=0$
$\implies x=\dfrac23,x=2$

are the critical points, and judging by the picture alone, you must have $b=\dfrac23$ and $a=2$ . (You might want to verify with the derivative test in case that's expected.)

Then the shaded region has area

$\displaystyle\int_0^2x(x-2)^2\,\mathrm dx=\dfrac43$

I'll leave the details to you.

Now, for part (iv), you're asked to find the minimum of $\dfrac{\mathrm dy}{\mathrm dx}=y'$ , which entails first finding the second derivative:

$y'=3x^2-8x+4$
$\implies y''=6x-8$

setting equal to 0 and finding the critical point:

$6x-8=0\implies x=\dfrac86=\dfrac43$

This is to say the minimum value of $\dfrac{\mathrm dy}{\mathrm dx}$ *occurs when $x=\dfrac43$ *, but this is not necessarily the same as saying that $\dfrac43$ is the actual minimum value.

The minimum value of $\dfrac{\mathrm dy}{\mathrm dx}$ is obtained by evaluating the derivative at this critical point:

$m=\dfrac{\mathrm dy}{\mathrm dx}\bigg|_{x=4/3}=3\left(\dfrac43\right)^2-8\left(\dfrac43\right)+4=-\dfrac43$

4 0

3 years ago

A coat was marked down by 60 percent from its original price of $99. Including an 8 percent sales tax, what is the final cost of

Neporo4naja [7]

99.00 - 60% = 39.60 + 8% = 42.77

3 0

3 years ago

Read 2 more answers

Find the equation of a circle that has a diameter with endpoints located at (7,3) and (7,-5)?

Julli [10]

For the area it would be 50.27 but for the circumference; it would be <span>25.13 I hope this helped ^^</span>

5 0

3 years ago