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denis-greek [22]
1 year ago
9

I need help with this practice I attempted this practice previously and my attempt is in the picture

Mathematics
1 answer:
diamong [38]1 year ago
4 0

We will draw a sketch for the given triangle to find its area

The area of the triangle will be

A=\frac{1}{2}\times XY\times ZM

Since ZX = ZY = 7, then the triangle is isosceles

Then the height ZM will bisect the base XY

Then we can find ZM by using Pythagoras Theorem

\begin{gathered} ZM=\sqrt[]{7^2-3^2} \\ ZM=\sqrt[]{49-9} \\ ZM=\sqrt[]{40} \\ ZM=2\sqrt[]{10} \end{gathered}

Since XY = 6, then

The area of the triangle is

\begin{gathered} A=\frac{1}{2}\times6\times2\sqrt[]{10} \\ A=6\sqrt[]{10}\text{ square units} \end{gathered}

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What is the distance between the following coordinates? Show work.
miss Akunina [59]

Answer:

\huge{ \boxed{ \bold{ \tt{2 \sqrt{26}}}  \:  \: \text{units} } }

<h3>♁ Question :</h3>
  • Find the distance between ( 5 , -6 ) and ( 3 ,4 ).

<h3>♁ Step by step explanation:</h3>

Let the points be A and B. Now,

  • Let, A ( 5 , -6 ) ⇢ ( x₁ , y₁ )
  • Let, B ( 3 , 4 ) ⇢ ( x₂ , y₂ )

Use the distance formula to determine the distance between A ( 5 , -6 ) and B ( 3 , 4 ).

\boxed{ \underline{ \sf{distance =  \sqrt{ {(x2 - x1)}^{2}  +  {(y2 - y1)}^{2} } }}}

Substitute the actual value of the points into the distance formula and then simplify.

\dashrightarrow{ \sf{ \sqrt{ {(3 - 5)}^{2} +  {(4 - ( - 6))}^{2}  } }}

Remember : The positive and negative integer are always subtracted but posses the sign of the bigger integer.

\dashrightarrow{ \sf{ \sqrt{ {( - 2)}^{2}  +  {(4 - ( - 6))}^{2} } }}

Remember : ( - ) × ( - ) = ( + )

\dashrightarrow{ \sf{ \sqrt{ {( - 2)}^{2} +  {(4 + 6)}^{2}  } }}

Add the numbers : 4 and 6

\longrightarrow{ \sf{ \sqrt{ {( - 2)}^{2}  +  {(10)}^{2} } }}

Evaluate the power

\longrightarrow{ \sf{ \sqrt{4 + 100}}}

Add the numbers : 4 and 100

\dashrightarrow{ \sf{ \sqrt{104}}}

\dashrightarrow{ \boxed{ \sf{2 \sqrt{26} }}} units

Therefore , The distance between ( 5 , -6 ) and ( 3 , 4 ) is \sf{2 \sqrt{26}} units .

And we're done!

Hope I helped!

Have a wonderful time ! シ

~TheAnimeGirl

5 0
3 years ago
Answer the question MATH
stiv31 [10]

Answer:

C, 108

Step-by-step explanation:

rectangles

8x3 = 24

8x4 = 32

8x5 = 40

triangles

3x4 = 12

you dont really have to divide by 2 since theres two triangles

24+32+40+12 = 108

6 0
4 years ago
Use the fundamental identities to <br> Find tan s if sin s=3/4 and s is in quadrant 2
OleMash [197]

Answer:

Cosine Formula

Thus, the cosine of angle α in a right triangle is equal to the adjacent side's length divided by the hypotenuse. To solve cos, simply enter the length of the adjacent and hypotenuse and solve.

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2 years ago
-4/5=2b<br> How do I solve for b in this problem?
Nimfa-mama [501]
-4/5=2b
since b is alrteady in the subject of the equation simply divide both side of the equation by the coefficient of b
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3 years ago
Find a solution to the initial value problem,<br> y″+12x=0, y(0)=2,y′(0)=−1.
levacccp [35]

Answer:

y = -2*x^3 - x + 2

Step-by-step explanation:

We want to solve the differential equation:

y'' + 12*x = 0

such that:

y(0) = 2

y'(0) = -1

We can rewrite our equation to:

y'' = -12x

if we integrate at both sides, we get:

\int {y''} \, dx  = y'=  \int {-12x} \, dx

Solving that integral we can find the value of y', so we will get:

y' = -12* (1/2)*x^2 + C = -6*x^2 + C

where C is the constant of integration.

Evaluating y' in x = 0 we get:

y'(0) = -6*0^2 + C = C

and for the initial value problem, we know that:

y'(0) = -1

then:

y'(0) = -1 = C

C = -1

So we have the equation:

y' = -6*x^2 - 1

Now we can integrate again, to get:

y = -6*(1/3)*x^3 - 1*x + K

y =  -2*x^3 - x + K

Where K is the constant of integration.

Evaluating or function in x = 0 we get:

y(0) = -2*0^3 - 0 + K

y(0) = K

And by the initial value, we know that: y(0) = 2

Then:

y(0) = 2 = K

K = 2

The function is:

y = -2*x^3 - x + 2

4 0
3 years ago
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