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3 years ago

Help mefind the vlie of the n's n+n=8 1/2

Mathematics

1 answer:

lora16 [44]3 years ago

8 0

2n = 8 1/2

N = 4 1/4 (4.25)

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Solve the following problems. Remember that all reasoning must be explained, and all steps of math work must be shown!

chubhunter [2.5K]

<h2>Question 1:</h2>

<h3>How to solve Part A</h3>

Since the diagram shows a right triangle and gives out some measurements of the sides, you can use the Pythagorean Theorem ( $a^{2}+b^{2}=c^{2}$ ) to find the length of MP.

<u>Given:</u>

hypotenuse = 7 – because the radius is 7 units and it can be used as the hypotenuse

One side of the triangle = 4 – because the length of PO is 4 units and it can be used as one sides of the triangle

<u>To Solve:</u>

Since we know two values of the triangle, you can use the Pythagorean Theorem ( $a^{2}+b^{2}=c^{2}$ ) to find what the other value is. The variables <em>c</em> and <em>b</em> in the formula are already given, and we need to find what <em>a</em> is to find the length of MP. So, plug the given values of <em>c</em> and <em>b</em> into the formula:

$a^{2}+4^{2}=7^{2}$

Simplify:

$a^{2}+16=49$

Subtract 16 from both sides of equation:

$a^{2}=33$

Do the square root of both sides to isolate the variable <em>a</em>:

$\sqrt{a^{2}}=\sqrt{33}$

Evaluate:

$a=\sqrt{33}$ Which simplifies to $a= 5.744562646....$ and round it to the nearest tenth to finally get $a=5.7$

Answer: The length of MP is 5.7 units

<h3>How to solve Part A</h3>

If you look at the diagram in the question, you can see that the length of MN is twice the length of MP. So multiply the length of MP, which is 5.7, by 2:

5.7 · 2 = 11.4

Answer: The length of MN is 11.4 units

___________________________________________________________

<h2>Question 2:</h2>

<h3>How to solve Part A</h3>

Look at the <u>first image</u> below ↓

<u>Description of the first image:</u>

The first image shows a drawn image of the circle with labeled parts and measurements from the infomation given from the question.

The diameter is 26cm, which means that the radius is 13cm because the radius half of the diameter in a circle. The very top line is the surface of the water where it’s filled and it has a length of 20cm. Side <em>b</em> of the triangle is 10cm becuase it’s half the length of the the surface of the water bowl. The hypotenuse (side <em>c</em>) is 13cm because it’s the length of the radius. And side <em>a </em>is what what we need to find because it’s the distance between the surface of the water and the center of the bowl. But after when you find the length of side <em>a</em>, you need to add the length of the radius (which is 13cm) to the length of side <em>a</em> because that’s the rest of the length/depth of the water that’s filled in the bowl. The length of the radius is included with the depth of the water.

<u>Total information given:</u>

⇒ radius = 13cm – because the radius is half of the diameter, and the diameter has a length of 26cm

⇒ length of the surface where the water’s filled = 20cm

⇒ hypotenuse (side <em>c</em>) = 13cm – because it’s the length of the radius

⇒ side <em>b</em> = 10cm – because it’s half the length of the surface where the water’s filled

<u>To Solve</u>

Since there is a right triangle shown in the image, you can use the Pythagorean Theorem ( $a^{2}+b^{2}=c^{2}$ ) to find the missing length of side <em>a</em>. Plug the given values of <em>c</em> and <em>b</em> into the formula:

$a^{2}+10^{2}=13^{2}$

Simplify:

$a^{2}+100=169$

Subtract 100 from both sides of the equation:

$a^{2}=69$

Do the square root of both sides to isolate the variable <em>a</em>:

$\sqrt{a^{2}}=\sqrt{69}$

Evaluate:

$a=\sqrt{69}$ Which simplifies to $a= 8.30662386....$ and round it to the nearest tenth to finally get $a$ ≈ $8.3$

This means that 8.3 is just the length of side <em>a</em>, which is the length between the surface of where the water’s filled and the center of the circle. But, now you need to add 13 to 8.3 to find the total depth of the water because the length of the radius is included with the depth of the water.

So,

13 + 8.3 = 21.3

Answer: The depth of the water is approximately 21.3 cm

<h3>How to solve Part B (is in the image below)</h3>

<em />

<em>Sorry for the very, VERY long explanation and the long solving process, but I really, really hope you understand and that this helps with your question! </em>:)

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3 years ago

The following histogram shows the number of items sold at a grocery store at various prices:

monitta

Answer:

Step-by-step explanation:

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3 years ago

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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:

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

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