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cluponka [151]
3 years ago
10

A recipe requires 3/8 cup of sugar for each cup of flour used if a baker uses 10 cups of flour what is the total amount of sugar

that will be needed
Mathematics
2 answers:
posledela3 years ago
8 0
Hello there!
If the baker uses 3/8 cup of sugar for every cup of flour and if he uses 10 cups of flours, then you have to multiply 10/1 with 3/8
10*3/8 = 30/8. 30/8 is converted into 3 & 6/8.
The baker will need 3 & 3/4 cups of sugar.
Gnoma [55]3 years ago
3 0
Multiply 10(3/8)
You get 3.75 or 3&3/4 cups of sugar if 10 cups of flour are used.
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Determine the center and radius of the following circle equation:
ioda

Answer:

C(5,-9) and r=8.

Step-by-step explanation:

C(p, q) - center

r=radius

k:x^2 +y^2 +dx+ey+f=0

x^2+y^2 - 10x+18y+42=0

d=-10, e=18, f=42

p=-d/2=-(-10)/2=10/2=5

q=-e/2=-18/2=-9

r^2 =p^2+q^2-f

r^2 =5^2 +(-9)^2 - 42

r^2=25+81-42

r^2 =106-42

r^2 =64

r=sqrt(64)

r=8

C(p,q)=C(5,-9) r=8

5 0
3 years ago
What is the equation of the line parallel to 3x + 5y = 11 that passes through the point (15, 4)?
soldier1979 [14.2K]

Answer:

The answer is

<h2>y =  -  \frac{3}{5} x + 13</h2>

Step-by-step explanation:

Equation of a line is y = mx + c

where

m is the slope

c is the y intercept

To find the equation of the parallel line we must first find the slope of the original line

The original line is 3x + 5y = 11

We must first write the equation in the general equation above

So we have

5y = - 3x + 11

Divide both sides by 5

<h3>y =  -  \frac{ 3}{5} x +  \frac{11}{5}</h3>

Comparing with the general equation above

Slope = - 3/5

Since the lines are parallel their slope are also the same

Slope of parallel line = - 3/5

So the equation of the line using point

(15, 4) and slope - 3/5 is

<h3>y - 4 =  -  \frac{3}{5} (x - 15) \\y - 4 =  -  \frac{3}{5} x + 9 \\ y =  -  \frac{3}{5} x + 9 + 4</h3>

We have the final answer as

<h3>y =  -  \frac{3}{5} x + 13</h3>

Hope this helps you

8 0
3 years ago
Passes through the points (−4, 2) and (12, 6)?
svlad2 [7]

<u>I'll assume you need to find the equation of the line that passes through those points.</u>

Answer:

\displaystyle y-2=\frac{1}{4}(x+4)

Step-by-step explanation:

<u>Equation of a Line:</u>

The equation of a line passing through points (x1,y1) and (x2,y2) can be found as follows:

\displaystyle y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)

The given points are (-4,2) and (12,6), thus:

\displaystyle y-2=\frac{6-2}{12+4}(x+4)

Operating:

\displaystyle y-2=\frac{4}{16}(x+4)

Simplifying, the equation in point-slope form is:

\mathbf{\displaystyle y-2=\frac{1}{4}(x+4)}

3 0
2 years ago
I need help and can you please give me the answers as well appreciate it
EastWind [94]

I honestly need help with this

5 0
3 years ago
Solve the equation. Then check your solution
kondaur [170]

Answer:

  • Option <u>B </u>is correct i.e. <u>2</u><u>1</u>

Step-by-step explanation:

In the question we're provided with an equation that is :

  • v/7 = 3

And we are asked to find the solution for the equation .

<u>Solution</u><u> </u><u>:</u><u> </u><u>-</u>

<u>\longrightarrow \:  \frac{v}{7}  = 3</u>

Multiplying by 7 on both sides :

\longrightarrow \: \frac{v}{ \cancel{7} }\times  \cancel{7}  = 3 \times 7

On further calculations , we get :

\longrightarrow \:  \blue{\boxed{v = 21}}

  • <u>Therefore</u><u> </u><u>,</u><u> </u><u>solution</u><u> </u><u>for</u><u> equation</u><u> </u><u>is </u><u>2</u><u>1</u><u> </u><u>.</u><u>That </u><u>means</u><u> </u><u>option </u><u>B </u><u>is </u><u>the </u><u>correct</u><u> answer</u><u>.</u>

<u>Verifying</u><u> </u><u>:</u>

We are verifying our answer by substituting value of v in the equation given in question :

\longrightarrow \: \frac{v}{7}  = 3

Putting value of v :

\longrightarrow \:  \cancel{\frac{21}{7}}  = 3

By dividing 21 with 7 , we get :

\longrightarrow \:3 = 3

\longrightarrow \: L.H.S=R.H.S

\longrightarrow \: Hence , \: Verified.

  • <u>Therefore</u><u> </u><u>,</u><u> </u><u>our </u><u>answer</u><u> is</u><u> valid</u><u> </u><u>.</u>

<h2><u>#</u><u>K</u><u>e</u><u>e</u><u>p</u><u> </u><u>Learning</u></h2>
3 0
2 years ago
Read 2 more answers
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