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3 years ago

Can somone pls help me .. i will mark u brainliest !

Mathematics

1 answer:

frutty [35]3 years ago

6 0

Answer:

1. D

2. A

3. B

Step-by-step explanation:

1. Add exponents

2. Multiply exponents to outside exponent and to the 16

3. Distribute exponents

You might be interested in

A cube-shaped art sculpture has a volume of 64 cubic

trasher [3.6K]

Answer:

4 ft

Step-by-step explanation:

The volume of a cube of side length s is V = s^3. Solving for s^3, we get:

s^3 = V, and so s = ∛V

In this case, with V = 64 ft³, the desired side length is

s = ∛(64 ft³) = 4 ft

4 0

3 years ago

Name a point that is 20 units away from (5,-2)

Fudgin [204]

Answer:

Distance between both points is 20units.

Applying the Distance between Points Formula

d= √(y₂-y₁)²+(x₂-x₁)²

x₁= 5 y₁=–2

x₂= ? y₂= ?

20 = √ [y -(-2)]² + (x - 5)²

Taking the square of both sides to eliminate the square root on the right.

20² = ( y + 2)² + (x - 5)²

400 = y² + 4y + 4 + x² – 10x + 25.

400 = x² + y² – 10x + 4y + 29.

x² + y² – 10x + 4y = 400 – 29

x² + y² – 10x + 4y = 371--------------eqn 1.

We called that eqn 1 cause we need another equation to solve this.

To solve an Equation with 2 variables... You need 2 equation.

To solve that of 3 variables.... you need 3 equations(Basic Math Rules).

We're dealing with 2 variables in this case(x, y).

Now

We'd need to think in order to create another equation that'll satisfy the distance or point.

We know that the distance between both Points(known and unknown) is 20.

Half that distance is 10.

The distance of 10 will occur at the center of these 2 points.

So lets apply the formula for Mid point to get the coordinate of the center.

Midpoint

let p be the x position and q be the y position of the midpoint

p = (x₁ + x₂)/2. q = (y₁ + y₂)/2

Starting from the know point (5 , -2)

x₁=5 y₁=–2

p = (5 + x)/2. q = (–2 + y)/2

p = (x + 5)/2 q = (y–2)/2.

This is the coordinate of the Midpoint.

Now We know that the distance between (5, -2) and [(x + 5)/2 , (y–2)/2] is 10(midpoint).

Applying the distance between points again.

d = √(y₂ - y₁)² + (x₂ - x₁)²

x₁= 5 y₁= -2

x₂= (x + 5)/2 y₂= (y–2)/2

10 = √[(y–2)/2 -(-2)]² + [ (x+5)/2 – 5)]²

10 = √[(y–2)/2 + 2]² + [(x+5)/2 – 5)]²

Squaring both sides to remove square root on the right and also Taking the LCM in each bracket

We have

10² = [ (y + 2)/2]² + [ (x–5)/2]²

Distributing the "2" on the denominator to each individual number in the parentheses.

100 = (y/2 + 1)² + (x/2 – 5/2)²

100 = y²/4 + y + 1 + x²/4 –5x/2 + 25/4.

Multiply through by 4 to clear fractions.

400 = y² + 4y + 4 + x² – 10x + 25

Rearranging

x² + y² – 10x + 4y = 371.

Hmm...

Equation 1 and 2 Are the Same.

Something is Wrong!

Pls Recheck the Question

I do believe that One value of the unknown point should be given.

I maybe be wrong somewhere else too.

Pls correct in the comment section if you spot it.

7 0

3 years ago

I forgot how to do this.... can somebody help?

slamgirl [31]

$\bf \qquad \qquad \qquad \qquad \textit{function transformations}
\\ \quad \\\\

\begin{array}{rllll} 
% left side templates
f(x)=&{{ A}}({{ B}}x+{{ C}})+{{ D}}
\\ \quad \\
y=&{{ A}}({{ B}}x+{{ C}})+{{ D}}
\\ \quad \\
f(x)=&{{ A}}\sqrt{{{ B}}x+{{ C}}}+{{ D}}
\\ \quad \\
f(x)=&{{ A}}(\mathbb{R})^{{{ B}}x+{{ C}}}+{{ D}}
\\ \quad \\
f(x)=&{{ A}} sin\left({{ B }}x+{{ C}} \right)+{{ D}}
\end{array}

$

$\bf \begin{array}{llll}
% right side info
\bullet \textit{ stretches or shrinks horizontally by } {{ A}}\cdot {{ B}}\\\\
\bullet \textit{ flips it upside-down if }{{ A}}\textit{ is negative}
\\\\
\bullet \textit{ horizontal shift by }\frac{{{ C}}}{{{ B}}}\\
\qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is negative, to the right}\\\\
\qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is positive, to the left}\\\\
\end{array}$

$\bf \begin{array}{llll}


\bullet \textit{ vertical shift by }{{ D}}\\
\qquad if\ {{ D}}\textit{ is negative, downwards}\\\\
\qquad if\ {{ D}}\textit{ is positive, upwards}\\\\
\bullet \textit{ period of }\frac{2\pi }{{{ B}}}
\end{array}$

now, with that template in mind, let's see

3 units to the right, that means C/B = -3 so hmm C = -3 and B = 1 will do, -3/1 = -3

vertical stretch by 2, so A = 2
reflected over the x-axis, so that means is flipped upside-down, so A = -2 then

and shifted down by 3, do D = -3

$\bf f(x)={{ A}}(\mathbb{R})^{{{ B}}x+{{ C}}}+{{ D}}\implies f(x)={{ -2}}(\mathbb{R})^{{{ 1}}x-{{ 3}}}-{{3}}\\\\\\ f(x)={{ -2}}(3)^{{{ }}x-{{ 3}}}-{{3}}$

3 0

4 years ago

There are 140 members in a community sports program. Sixty-five percent of these members play soccer.

Aleks [24]

75 people do not play soccer

6 0

3 years ago

Read 2 more answers

A ball is thrown in the air that is 48 feet above the ground with an initial vertical velocity of 32 feet per second. The height

anastassius [24]

Answer:

h(t) = -16(t - 1)² + 64

Step-by-step explanation:

The function that would best to use to identify the maximum height of the ball is the vertex form of the parabola.

The standard form of a quadratic function is

y = ax² + bx + c

The vertex form is

y = a(x - h)² + k

where (h, k) is the vertex of the parabola.

h = -b/(2a) and k = f(h)

In your equation, h(t) = -16t^2 + 32t + 48

a = -16; b = 32; c = 48

Calculate h

h = -32/[2(-16)]

h = (-32)/(-32)

h = 1

Calculate k

k = -16(1)² + 32×1 + 48

k = -16 + 32 + 48

k = 64

So, h = 1, k = 64, a = -16

The vertex form of the equation is h(t) = -16(t - 1)² + 64.

The graph below shows h(t) with the ball at its maximum height of 64 ft one second after being thrown.

7 0

3 years ago