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3 years ago

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Mandy has a 10.2-ounce container of mustard. She uses 130 of the mustard to put on her hot dog each time. How many ounces of mus

tard are left after she has eaten 6 hot dogs?
Record your answer and fill in the bubbles on your answer document. Be sure to use the correct place value.

2 answers:

Marianna [84]3 years ago

7 0

Answer:

x

Step-by-step explanation:

Triss [41]3 years ago

7 0

This is from a test isnt it? Is this the snapshot i had exactly this.......CHEATER O-o

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Points A, B, and C, form a triangle. The distance between point A and point B is 15 yards. The distance between point B and poin

Natasha_Volkova [10]

The total distance if Pete walks from A to B to C is 40 yards, the direct distance from A to C is more than 10 yards, and an inequality w < 30

<h3>What is inequality?</h3>

It is defined as the expression in mathematics in which both sides are not equal they have mathematical signs either less than or greater than known as inequality.

We have:

Points A, B, and C, form a triangle. The distance between point A and point B is 15 yards. The distance between point B and point C is 25 yards.

Total distance if Pete walks from A to B to C:

= 15 + 25 = 40 yards

As we know in the triangle the sum of the two sides is greater than the third side.

25 + 15 > AC

The direct distance from A to C is more than 10(25-15) yards.

The inequality w < 30 represents the distance, w, that Pete might save by taking the direct path.

Thus, the total distance if Pete walks from A to B to C is 40 yards, the direct distance from A to C is more than 10 yards, and an inequality w < 30

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2 years ago

Find the arc length of the given curve between the specified points. x = y^4/16 + 1/2y^2 from (9/16), 1) to (9/8, 2).

lutik1710 [3]

Answer:

The arc length is $\dfrac{21}{16}$

Step-by-step explanation:

Given that,

The given curve between the specified points is

$x=\dfrac{y^4}{16}+\dfrac{1}{2y^2}$

The points from $(\dfrac{9}{16},1)$ to $(\dfrac{9}{8},2)$

We need to calculate the value of $\dfrac{dx}{dy}$

Using given equation

$x=\dfrac{y^4}{16}+\dfrac{1}{2y^2}$

On differentiating w.r.to y

$\dfrac{dx}{dy}=\dfrac{d}{dy}(\dfrac{y^2}{16}+\dfrac{1}{2y^2})$

$\dfrac{dx}{dy}=\dfrac{1}{16}\dfrac{d}{dy}(y^4)+\dfrac{1}{2}\dfrac{d}{dy}(y^{-2})$

$\dfrac{dx}{dy}=\dfrac{1}{16}(4y^{3})+\dfrac{1}{2}(-2y^{-3})$

$\dfrac{dx}{dy}=\dfrac{y^3}{4}-y^{-3}$

We need to calculate the arc length

Using formula of arc length

$L=\int_{a}^{b}{\sqrt{1+(\dfrac{dx}{dy})^2}dy}$

Put the value into the formula

$L=\int_{1}^{2}{\sqrt{1+(\dfrac{y^3}{4}-y^{-3})^2}dy}$

$L=\int_{1}^{2}{\sqrt{1+(\dfrac{y^3}{4})^2+(y^{-3})^2-2\times\dfrac{y^3}{4}\times y^{-3}}dy}$

$L=\int_{1}^{2}{\sqrt{1+(\dfrac{y^3}{4})^2+(y^{-3})^2-\dfrac{1}{2}}dy}$

$L=\int_{1}^{2}{\sqrt{(\dfrac{y^3}{4})^2+(y^{-3})^2+\dfrac{1}{2}}dy}$

$L=\int_{1}^{2}{\sqrt{(\dfrac{y^3}{4}+y^{-3})^2}dy}$

$L= \int_{1}^{2}{(\dfrac{y^3}{4}+y^{-3})dy}$

$L=(\dfrac{y^{3+1}}{4\times4}+\dfrac{y^{-3+1}}{-3+1})_{1}^{2}$

$L=(\dfrac{y^4}{16}+\dfrac{y^{-2}}{-2})_{1}^{2}$

Put the limits

$L=(\dfrac{2^4}{16}+\dfrac{2^{-2}}{-2}-\dfrac{1^4}{16}-\dfrac{(1)^{-2}}{-2})$

$L=\dfrac{21}{16}$

Hence, The arc length is $\dfrac{21}{16}$

6 0

3 years ago