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3 years ago

Find the height of ABD

Mathematics

1 answer:

Mila [183]3 years ago

7 0

Answer:

$Height=\boxed{10\sqrt{2}}$

$Area =\boxed{165\sqrt{2}}$

Step-by-step explanation:

By geometric mean theorem:

$AC=\sqrt{25\times 8}$

$AC=\sqrt{5^2 \times 2^2 \times 2}$

$AC=5\times 2\sqrt{2}$

$AC=10\sqrt{2}$

So,

$Height=\boxed{10\sqrt{2}}$

$Area = \frac{1}{2} \times base\times height$

$\therefore Area = \frac{1}{2} \times (25+8)\times 10\sqrt{2}$

$\therefore Area =33\times 5\sqrt{2}$

$Area =\boxed{165\sqrt{2}}$

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Kaia has $525 in her savings account. The next month the balance is $635. The balance after the third month is $690. Write a fun

kvasek [131]

Answer:jm i the the the the 539

Step-by-step explanation:

8 0

4 years ago

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

1 year ago

Find the measure of the angle indicated (x). Round to the nearest tenth Show your work to support your answer

JulijaS [17]

Answer:

Karen, this type of question is such a tricky question, there is a lot of things that you just have to remember for these. Also, this type of question is asked a lot on brainly, you are in good company :P This also makes me wonder if math teachers are just REALLY BAD at teaching this.. hmmm Something is up with this type of problem b/c I have answer 100s of these type on here.. they are tricky. soooo...

Step-by-step explanation:

1st ,

Use SOH CAH TOA to recall how the trig functions fit on a triangle

SOH: Sin(Ф)= Opp / Hyp

CAH: Cos(Ф)= Adj / Hyp

TOA: Tan(Ф) = Opp / Adj

with the above info, we can solve your triangle :P not meant to sound "wrong" .. anyway

look for the info you are given and what you want to find... and match that to one of the 3 formulas. after you've done this a few 100 times.. you'll just remember the above formulas. :D

sooo I see that we want to find that angle ∅ and we know two of the sides and that this is a right triangle.. which means we can use the above formulas on this triangle. So we know the adjacent side and the opposite side, to the angle, soo lets use TOA

Tan(∅) = Opp / Adj

Tan(∅) = 4 / 13

now the algebra gets a little tricky here.. but it's just using the inverse function of the Tan to find the angle .. it's just going backwards with the function is all.. nothing to see here... move along.. :D

take the inverse Tan function ( arcTan) on both sides... sooo

∅ = arcTan( 4 /13 ) ( this will find the angle, but make sure your calculator is set to degrees.. and not radians )

∅ = 17.1027...°

So, now we know that angle ∅

SOH CAH TOA is super helpful for this kind of triangle problem :)

7 0

3 years ago

Write an equation in slope-intercept form for the line with slope and -intercept .

Mazyrski [523]

Answer:

y = $\frac{2}{5}$ x - 9

Step-by-step explanation:

6 0

3 years ago

What is the distance between the points (5,1) and (-3,-5)?<br>

Charra [1.4K]

Answer

$\boxed{10 \: \: units}$

Step by step explanation

Let the points be A and B

A ( 5 , 1 ) ⇒ ( x₁ , y₁ )

B ( -3 , -5 )⇒ ( x₂ , y₂ )

Now, let's find the distance between theses two points:

Distance = $\mathsf{ \sqrt{ {(x2 - x1)}^{2} + {(y2 - y1)}^{2} } }$

Plug the values

$\mathsf{ \sqrt{ {( - 3 - 5)}^{2} + {( - 5 - 1)}^{2} } }$

Calculate

$\mathsf{ \sqrt{ { ( - 8)}^{2} + {( - 6)}^{2} } }$

Evaluate the power

$\mathsf{ \sqrt{64 + 36} }$

Add the numbers

$\mathsf{ \sqrt{100} }$

Write the number in exponential form with a base of 10

$\mathsf {\sqrt{ {10}^{2} } }$

Reduce the index of the radical and exponent with 2

$\mathsf{10 \: units}$

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The distance formula is used to determine the distance ( d ) between two points. If the co-ordinates of the two points are ( x₁ , y₁) and ( x₂ , y₂ ) , the distance equals the square root of x₂ - x₁ squared + y₂ - y₁ squared.

Hope I helped!

Best regards!

6 0

3 years ago