Ask question

Ask question

All categories

2 years ago

10

A bag contains blue marbles and red marbles, 107 in total. The number of blue marbles is 7 more than 4 times the number of red m

arbles. How many blue marbles are there?

1 answer:

Alecsey [184]2 years ago

6 0

Answer: blue= 93.625

Step-by-step explanation:

Let the red ball be y

Four times of red is 4y

And blue is seven times of red which means 7(4y)

4y+7(4y)=107

Open the bracket

4y+28y=107

32y=107

y=107/32

Make y the subject of formula

y=3.34

Therefore,blue is

7(4(3.34))

Blue= 93.625

You might be interested in

I get paid $40 a month. If i want 5 less each month, how much should i get paid every week?

agasfer [191]

Answer:

$5 each week

Step-by-step explanation:

$40 a month

$10 a week

$10-$5 is $5 each week

4 0

2 years ago

Read 2 more answers

Point D is in the interior of ∠ABC. What is m∠DBC?

Rudiy27

37.5 degrees

we know angled like this always have 180 degrees total between the 2. which means we can set up a little equation to solve for x. The equation is (3x+22)+(x-4)= 180. Initially we can take the solid numbers (22 and -4) and get rid of them by taking them to the 180. 180- 22 is 158. 158 + 4 is 162. now we have 3x + x = 162. we combine like terms. means we end up with 4x =162. We now divide by 4 which means we get 40.5. Now we know x is 40.5. Now dbc is just 40.5- 4 which is 37.5.

3 0

2 years ago

At Southside Middle School, 20% of the students play an instrument. There are 305 students. How many students play an instrument

RoseWind [281]

Answer:

61 Students.

Step-by-step explanation:

So you need to find 20% of 305. You can do this by dividing by 5.

305 divided by 5 equals 61.

If you have questions, ask please :)

4 0

2 years ago

Read 2 more answers

Find the area of the following ellipse. a = 4.5 m; b = 5.5 m

LenaWriter [7]

Answer:

Area of ellipse is ≈ 77.78 $m^{2}$

Step-by-step explanation:

Given the dimension of ellipse

a = 4.5 m

b = 5.5 m

Now, Area of ellipse = $\pi ab$

= $\frac{22}{7}\times 4.5\times 5.5$

= $\frac{5445}{70}$

= 77.78 $m^{2}$

Hence Area of ellipse will be 77.78 $m^{2}$

6 0

2 years ago

Read 2 more answers

F(3) = 8; f^ prime prime (3)=-4; g(3)=2,g^ prime (3)=-6 , find F(3) if F(x) = root(4, f(x) * g(x))

Marrrta [24]

Given:

$f(3)=8,f^{\prime}(3)=-4,g(3)=2,\text{ and }g^{\prime}(3)=-6$

Required:

$We\text{ need to find }F^{\prime}(3)\text{ if }F(x)=\sqrt[4]{f(x)g(x)}.$

Explanation:

Given equation is

$F(x)=\sqrt[4]{f(x)g(x)}.$ $F(x)=(f(x)g(x))^{\frac{1}{4}}$ $F(x)=f(x)^{\frac{1}{4}}g(x)^{\frac{1}{4}}$

Differentiate the given equation for x.

$Use\text{ }(uv)^{\prime}=uv^{\prime}+vu^{\prime}.\text{ Here u=}\sqrt[4]{f(x)}\text{ and v=}\sqrt[4]{g(x)}.$

$F^{\prime}(x)=f(x)^{\frac{1}{4}}(\frac{1}{4}g(x)^{\frac{1}{4}-1})g^{\prime}(x)+g(x)^{\frac{1}{4}}(\frac{1}{4}f(x)^{\frac{1}{4}-1})f^{\prime}(x)$ $=\frac{1}{4}f(x)^{\frac{1}{4}}g(x)^{\frac{1}{4}-\frac{1\times4}{4}}g^{\prime}(x)+\frac{1}{4}g(x)^{\frac{1}{4}}f(x)^{\frac{1}{1}-\frac{1\times4}{4}}f^{\prime}(x)$ $=\frac{1}{4}f(x)^{\frac{1}{4}}g(x)^{\frac{1-4}{4}}g^{\prime}(x)+\frac{1}{4}g(x)^{\frac{1}{4}}f(x)^{\frac{1-4}{4}}f^{\prime}(x)$ $F^{\prime}(x)=\frac{1}{4}f(x)^{\frac{1}{4}}g(x)^{\frac{-3}{4}}g^{\prime}(x)+\frac{1}{4}g(x)^{\frac{1}{4}}f(x)^{\frac{-3}{4}}f^{\prime}(x)$

Replace x=3 in the equation.

$F^{\prime}(3)=\frac{1}{4}f(3)^{\frac{1}{4}}g(3)^{\frac{-3}{4}}g^{\prime}(3)+\frac{1}{4}g(3)^{\frac{1}{4}}f(3)^{\frac{-3}{4}}f^{\prime}(3)$ $Substitute\text{ }f(3)=8,f^{\prime}(3)=-4,g(3)=2,\text{ and }g^{\prime}(3)=-6\text{ in the equation.}$ $F^{\prime}(3)=\frac{1}{4}(8)^{\frac{1}{4}}(2)^{\frac{-3}{4}}(-6)+\frac{1}{4}(2)^{\frac{1}{4}}(8)^{\frac{-3}{4}}(-4)$ $F^{\prime}(3)=\frac{-6}{4}(8)^{\frac{1}{4}}(2^3)^{\frac{-1}{4}}+\frac{-4}{4}(2)^{\frac{1}{4}}(8^3)^{\frac{-1}{4}}$ $F^{\prime}(3)=\frac{-3}{2}(8)^{\frac{1}{4}}(8)^{\frac{-1}{4}}-(2)^{\frac{1}{4}}(8^3)^{\frac{-1}{4}}$ $F^{\prime}(3)=\frac{-3}{2}\frac{\sqrt[4]{8}}{\sqrt[4]{8}}-\frac{\sqrt[4]{2}}{\sqrt[4]{8^3}}$ $F^{\prime}(3)=\frac{-3}{2}-\frac{\sqrt[4]{2}}{\sqrt[4]{(2)^9}}$ $F^{\prime}(3)=\frac{-3}{2}-\frac{\sqrt[4]{2}}{\sqrt[4]{(2)^4(2)^4}(2)}$ $F^{\prime}(3)=\frac{-3}{2}-\frac{\sqrt[4]{2}}{4\sqrt[4]{}(2)}$ $F^{\prime}(3)=\frac{-3}{2}-\frac{1}{4}$ $F^{\prime}(3)=\frac{-3\times2}{2\times2}-\frac{1}{4}$ $F^{\prime}(3)=\frac{-6-1}{4}$ $F^{\prime}(3)=\frac{-7}{4}$

Final answer:

$F^{\prime}(3)=\frac{-7}{4}$

8 0

9 months ago