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2 years ago

9

A caterer needs to buy 22 pounds of pasta to cater a wedding. At a local store, 8 pounds of pasta cost $12. How much will the ca

terer pay for the pasta there? work shown please.

1 answer:

kodGreya [7K]2 years ago

6 0

Answer:

1. 8/$12=21/$x then cross multiply.

2. 1/$1.5, we're looking for cost so $1 is more reasonable

3. $31.5

Step-by-step explanation:

12x 21/ 8 = 31.5

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9/4 = blank divided by 4<br><br> please help- what is blank

Mashcka [7]

Answer:

9

Step-by-step explanation:

You gave the answer right in your question

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3 years ago

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0.04 is 1/10 of what number

bulgar [2K]

If you would like to calculate 1/10 of what number is 0.04, you can do this using the following steps:

1/10 of what number is 0.04
1/10 * x = 0.04 /*10
x = 0.04 * 10
x = 0.4

Result: 1/10 of 0.4 is 0.04.

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3 years ago

please help :{ brainliest if you can get it right and I need you to prove why your answer is correct, thanksss

beks73 [17]

Answer:

The Value of $a=\frac{15}{4}$ .

Step-by-step explanation:

We have Named the figure please find the attachment for your reference.

Given:

PR = y

QR = a

RS = b

PS = z

PQ = x

QS = 15

∠P = 90°

∠R = 90°

∠Q = 60°

∠S = 30°

We need to find the Value of 'a'.

Solution:

Now we know that:

In Δ PQS

∠P = 90°

∠S = 30°

Now we know that;

$sin\ \theta = \frac{opposite\ side}{Hypotenuse}$

$sin \ S= \frac{PQ}{QS}$

Substituting the given values we get;

$sin\ 30\°=\frac{x}{15}$

Now we know that;

$sin\ 30\° = \frac12$

So we can say that;

$\frac{1}{2}=\frac{x}{15}\\\\x=\frac{15}{2}$

Now In Triangle PQR.

∠R = 90°

∠Q = 60°

So we can say that;

$Cos \theta = \frac{adjacent \ Side}{Hypotenuse}\\$

$Cos\ Q = \frac{QR}{PQ}$

Substituting the given values we get;

$cos 60\°= \frac{a}{x}$

Now we know that;

$cos 60\°= \frac12$

$x=\frac{15}{2}$

So substituting the values we get;

$\frac{1}{2}=\frac{a}{\frac{15}{2}}$

By Using Cross Multiplication we get;

$a= \frac{1}{2}\times\frac{15}{2}\\\\a=\frac{15}{4}$

Hence The Value of $a=\frac{15}{4}$ .

6 0

3 years ago

Brady made a scale drawing of a rectangular swimming pool on a coordinate grid. The points (-20, 25), (30, 25), (30, -10) and (-

djverab [1.8K]

Answer:

Length = 50 units

width = 35 units

Step-by-step explanation:

Let A, B, C and D be the corner of the pools.

Given:

The points of the corners are.

$A(x_{1}, y_{1}})=(-20, 25)$

$B(x_{2}, y_{2}})=(30, 25)$

$C(x_{3}, y_{3}})=(30, -10)$

$D(x_{4}, y_{4}})=(-20, -10)$

We need to find the dimension of the pools.

Solution:

Using distance formula of the two points.

$d(A,B)=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$ ----------(1)

For point AB

Substitute points A(30, 25) and B(30, 25) in above equation.

$AB=\sqrt{(30-(-20))^{2}+(25-25)^{2}}$

$AB=\sqrt{(30+20)^{2}}$

$AB=\sqrt{(50)^{2}$

AB = 50 units

Similarly for point BC

Substitute points B(-20, 25) and C(30, -10) in equation 1.

$d(B,C)=\sqrt{(x_{3}-x_{2})^{2}+(y_{3}-y_{2})^{2}}$

$BC=\sqrt{(30-30)^{2}+((-10)-25)^{2}}$

$BC=\sqrt{(-35)^{2}}$

BC = 35 units

Similarly for point DC

Substitute points D(-20, -10) and C(30, -10) in equation 1.

$d(D,C)=\sqrt{(x_{3}-x_{4})^{2}+(y_{3}-y_{4})^{2}}$

$DC=\sqrt{(30-(-20))^{2}+(-10-(-10))^{2}}$

$DC=\sqrt{(30+20)^{2}}$

$DC=\sqrt{(50)^{2}}$

DC = 50 units

Similarly for segment AD

Substitute points A(-20, 25) and D(-20, -10) in equation 1.

$d(A,D)=\sqrt{(x_{4}-x_{1})^{2}+(y_{4}-y_{1})^{2}}$

$AD=\sqrt{(-20-(-20))^{2}+(-10-25)^{2}}$

$AD=\sqrt{(-20+20)^{2}+(-35)^{2}}$

$AD=\sqrt{(-35)^{2}}$

AD = 35 units

Therefore, the dimension of the rectangular swimming pool are.

Length = 50 units

width = 35 units

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3 years ago

What is the measure of the arc represented by the numbers 6 and 11 on a clock face?

Marat540 [252]

150; because 6-11 is almost 180, but it’s not quite there, so it’s a little less

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3 years ago

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