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3 years ago

What is 3=7(4 - 2u)-6u

Mathematics

2 answers:

Elodia [21]3 years ago

8 0

Answer:

$u = 5/4$

Step-by-step explanation:

Step 1: Distribute on the right side

$3 = 7(4 - 2u) - 6u$

$3 = (7 * 4) + (7 * -2u) - 6u$

$3 = 28 - 14u - 6u$

Step 2: Combine like terms

$3 = 28 + (-14u - 6u)$

$3 = 28 + -20u$

Step 3: Subtract 28 from both sides

$3 - 28 = 28 - 20u - 28$

$-25 = -20u$

Step 4: Divide both sides by -20

$-25 / -20 = -20u / -20$

$5/4 = u$

Answer: $u = 5/4$

alisha [4.7K]3 years ago

5 0

Answer:

u = 5/4

Step-by-step explanation:

to evaluate for the value of u we would simply open the bracket and then evaluate for the value of u by collecting the like terms together.

solution

3=7(4 - 2u)-6u

3 = 28- 14u - 6u

collect the like terms

3 + 14u + 6u = 28

20u = 28 - 3

20u = 25

divide both sides by the coefficient of u which is 20

20u/20 = 25/20

u = 5/4

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Answer:

9 cans of tomato soup

Step-by-step explanation:

We are told that only cans of tomato soup and cans of chicken soup are put on the shelf.

Now, we are told that the ratio of the number of cans of chicken noodle soup to the total number of cans on the shelf was 8 to 17.

In fraction, this is 8/17

Now, this implies that there are 8 cans of chicken noodle soup while there are a total of 17 cans on the shelf.

Thus,

Since only cans of chicken noodles and cans of tomato soup are on the shelf, it means;

Cans of tomato soup = 17 - 8 = 9

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3 years ago

Apply the distributive property to create an equivalent expression.

Amanda [17]

The distributive property: a(b + c) = ab + ac

(-7c + 8d)0.6 = (-7c)(0.6) + (8d)(0.6) = -4.2c + 4.8d

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3 years ago

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each blade on a windmill is in the shape of a triangle with a base of 7 feet and a height of 4 ft if there are four blades on th

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The total area of the blades is the area of a triangle multiplied by 4.

We have then:

$A = (4) * (1/2) * (b) * (h)$

Where,

b: base of the blades

h: height of the blades

Substituting values we have:

$A = (4) * (1/2) * (7) * (4)$

Doing the calculation we have that the area of the blades is given by:

$A = 56 feet ^ 2$

Answer:

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$A = 56 feet ^ 2$

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2 years ago

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To find 10, Beau found 32 = 9 and 42 = 16. He said that since 10 is between 9 and 16, 10 is between 3

AlexFokin [52]

Given that:

Consider it is $\sqrt{10}$ instead of 10 on two places.

$\sqrt{10}$ is between 3 and 4. So, Beau thinks a good estimate for $\sqrt{10}$ is = 3.5.

Solution:

To find $\sqrt{10}$ , Beau found 3² = 9 and 4² = 16.

He said that since 10 is between 9 and 16.

Since 10 is close to 9, therefore $\sqrt{10}$ must be close to 3. So, Beau's estimate is high.

Now,

$(3.1)^2=9.61$

$(3.2)^2=10.24$

Since, 10 lies between 9.61 and 10.24, therefore $\sqrt{10}$ must be lies between 3.1 and 3.2.

$\dfrac{3.1+3.2}{2}=3.15$

Therefore, the estimated value of $\sqrt{10}$ is 3.15.

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3 years ago

In the triangle pictured, let A, B, C be the angles at the three vertices, and let a,b,c be the sides opposite those angles. Acc

Troyanec [42]

Answer:

Step-by-step explanation:

(a)

Consider the following:

$A=\frac{\pi}{4}=45°\\\\B=\frac{\pi}{3}=60°$

Use sine rule,

$\frac{b}{a}=\frac{\sinB}{\sin A}
\\\\=\frac{\sin{\frac{\pi}{3}}
}{\sin{\frac{\pi}{4}}}\\\\=\frac{[\frac{\sqrt{3}}{2}]}{\frac{1}{\sqrt{2}}}\\\\=\frac{\sqrt{2}}{2}\times \frac{\sqrt{2}}{1}=\sqrt{\frac{3}{2}}$

Again consider,

$\frac{b}{a}=\frac{\sin{B}}{\sin{A}}
\\\\\sin{B}=\frac{b}{a}\times \sin{A}\\\\\sin{B}=\sqrt{\frac{3}{2}}\sin {A}\\\\B=\sin^{-1}[\sqrt{\frac{3}{2}}\sin{A}]$

Thus, the angle B is function of A is, $B=\sin^{-1}[\sqrt{\frac{3}{2}}\sin{A}]$

Now find $\frac{dB}{dA}$

Differentiate implicitly the function $\sin{B}=\sqrt{\frac{3}{2}}\sin{A}$ with respect to A to get,

$\cos {B}.\frac{dB}{dA}=\sqrt{\frac{3}{2}}\cos A\\\\\frac{dB}{dA}=\sqrt{\frac{3}{2}}.\frac{\cos A}{\cos B}$

When $A=\frac{\pi}{4},B=\frac{\pi}{3}$ , the value of $\frac{dB}{dA}$ is,

$\frac{dB}{dA}=\sqrt{\frac{3}{2}}.\frac{\cos {\frac{\pi}{4}}}{\cos {\frac{\pi}{3}}}\\\\=\sqrt{\frac{3}{2}}.\frac{\frac{1}{\sqrt{2}}}{\frac{1}{2}}\\\\=\sqrt{3}$

In general, the linear approximation at x= a is,

$f(x)=f'(x).(x-a)+f(a)$

Here the function $f(A)=B=\sin^{-1}[\sqrt{\frac{3}{2}}\sin{A}]$

At $A=\frac{\pi}{4}$

$f(\frac{\pi}{4})=B=\sin^{-1}[\sqrt{\frac{3}{2}}\sin{\frac{\pi}{4}}]\\\\=\sin^{-1}[\sqrt{\frac{3}{2}}.\frac{1}{\sqrt{2}}]\\\\\=\sin^{-1}(\frac{\sqrt{2}}{2})\\\\=\frac{\pi}{3}$

And,

$f'(A)=\frac{dB}{dA}=\sqrt{3}$ from part b

Therefore, the linear approximation at $A=\frac{\pi}{4}$ is,

$f(x)=f'(A).(x-A)+f(A)\\\\=f'(\frac{\pi}{4}).(x-\frac{\pi}{4})+f(\frac{\pi}{4})\\\\=\sqrt{3}.[x-\frac{\pi}{4}]+\frac{\pi}{3}$

Use part (c), when $A=46°$ , B is approximately,

$B=f(46°)=\sqrt{3}[46°-\frac{\pi}{4}]+\frac{\pi}{3}\\\\=\sqrt{3}(1°)+\frac{\pi}{3}\\\\=61.732°$

8 0

3 years ago