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4 years ago

What is seven-tenths _ _ _ _ _ _ _ pieces of the whole?

Mathematics

1 answer:

satela [25.4K]4 years ago

3 0

The answer of seven-tenths in whole form is 70 out of 100

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Help me please<br> Idk what to do

Ber [7]

Answer:

x=36

Step-by-step explanation:

First rewrite equation then multiply each side by 36. 9x-36=4x+144.

then move the variable 9x-4x+36=144. Next subtract 9x and 4x ...5x=144+36.... 5x=180. Lastly you divide 180÷5 this your answer x=36

4 0

3 years ago

Jonah and his brother want to earn at least $400 this month, so they rented a lawn mower to mow lawns. They plan to charge $25 p

Travka [436]

Answer:

20 lawns

Step-by-step explanation:

25x4=100(duh) (16)

100x4=400(duh)

m=20

4 0

3 years ago

128,955 round to the nearest ten thousand

Studentka2010 [4]

Answer:

128,955 rounds to 130000 i believe

Step-by-step explanation:

4 0

3 years ago

Howdy!

posledela

Answer:

A) Central angle has same measure as intercepted arc.

mCE = mCD + mDE = 20° + 90° = 110°

B) Opposite angles of cyclic quadrilateral are supplementary.

mRL = 2*m∠PQR - mPL = 2*74° - 72° = 76°
m∠QPL = (1/2)mQRL = (1/2)(90° + 76°) = 83°
m∠QRL = 180° - m∠QPL = 180° - 83° = 97°
mQP = 360° - (90° + 76° + 72°) = 122°

m∠MLN = m∠MRN as same arc MN is intercepted
m∠LMN is right angle as opposite side is diameter.
∠MNL is complementary with ∠MLN which is same as ∠MRN
m∠MNL = 90° - 47° = 43°

D) Tangent secant angle is half of the intercepted arc.

<em>It seems wrong. Should be mQP instead of mQR</em>

mQP = 2*m∠RQP = 2*74° = 148°

8 0

3 years ago

A box with a square base and open top must have a volume of 157216 cm3. We wish to find the dimensions of the box that minimize

shepuryov [24]

Answer:

Base Length of 68cm
Height of 34 cm.

Step-by-step explanation:

Given a box with a square base and an open top which must have a volume of 157216 cubic centimetre. We want to minimize the amount of material used.

Step 1:

Let the side length of the base =x

Let the height of the box =h

Since the box has a square base

Volume $=x^2h=157216$

$h=\dfrac{157216}{x^2}$

Surface Area of the box = Base Area + Area of 4 sides

$A(x,h)=x^2+4xh\\$Substitute h=\dfrac{157216}{x^2}\\A(x)=x^2+4x\left(\dfrac{157216}{x^2}\right)\\A(x)=\dfrac{x^3+628864}{x}$

Step 2: Find the derivative of A(x)

$If\:A(x)=\dfrac{x^3+628864}{x}\\A'(x)=\dfrac{2x^3-628864}{x^2}$

Step 3: Set A'(x)=0 and solve for x

$A'(x)=\dfrac{2x^3-628864}{x^2}=0\\2x^3-628864=0\\2x^3=628864\\x^3=314432\\x=\sqrt[3]{314432}\\ x=68$

Step 4: Verify that x=68 is a minimum value

We use the second derivative test

$If\:A(x)=\dfrac{x^3+628864}{x}\\A''(x)=\dfrac{2x^3+1257728}{x^3}\\$When x=68\\A''(x)=6$

Since the second derivative is positive at x=68, then it is a minimum point.

Recall:

$h=\dfrac{157216}{x^2}\\h=\dfrac{157216}{68^2}=34$

Therefore, the dimensions that minimizes the box surface area are:

Base Length of 68cm
Height of 34 cm.

3 0

3 years ago