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2 years ago

I need help with these questions plz help

Mathematics

1 answer:

stepladder [879]2 years ago

6 0

Answer:

1.A=bh/2

2A=bh

2A/b=h

answer check

2.f(x)=21-6x f(5)=21-6(5)

g(x)=2x/5-11 f(5)=21-30

21-6x=2x/5-11 f(5)=-9

21=32x/5-11 g(5)=2(5)/5-11

(5)32=32x/5(5) g(5)= 10/5-11

160=32x g(5)=2-11

5=x g(5)=-9

3. f(7)=3(7)-7

f(7)= 21-7

f(7)= 14

Step-by-step explanation:

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Make denominator same first.

1/2 - 1/6

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(3-1) / 6

2/6

Now simplify

1/3

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3 years ago

Evaluate 3x2 - 2x when x = 5

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Answer:

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(HURRY! I'M BEING TIMED)Write the partial fraction decomposition of the rational expression.

Aleonysh [2.5K]

Answer:

The partial fraction decomposition is $\frac{- 4 x^{2} + 13 x - 12}{\left(x + 1\right)^{2} \left(x + 2\right)}=\frac{50}{x + 1}+\frac{-29}{\left(x + 1\right)^{2}}+\frac{-54}{x + 2}$ .

Step-by-step explanation:

Partial-fraction decomposition is the process of starting with the simplified answer and taking it back apart, of "decomposing" the final expression into its initial polynomial fractions.

To find the partial fraction decomposition of $\frac{- 4 x^{2} + 13 x - 12}{\left(x + 1\right)^{2} \left(x + 2\right)}$ :

First, the form of the partial fraction decomposition is

$\frac{- 4 x^{2} + 13 x - 12}{\left(x + 1\right)^{2} \left(x + 2\right)}=\frac{A}{x + 1}+\frac{B}{\left(x + 1\right)^{2}}+\frac{C}{x + 2}$

Write the right-hand side as a single fraction:

$\frac{- 4 x^{2} + 13 x - 12}{\left(x + 1\right)^{2} \left(x + 2\right)}=\frac{\left(x + 1\right)^{2} C + \left(x + 1\right) \left(x + 2\right) A + \left(x + 2\right) B}{\left(x + 1\right)^{2} \left(x + 2\right)}$

The denominators are equal, so we require the equality of the numerators:

$- 4 x^{2} + 13 x - 12=\left(x + 1\right)^{2} C + \left(x + 1\right) \left(x + 2\right) A + \left(x + 2\right) B$

Expand the right-hand side:

$- 4 x^{2} + 13 x - 12=x^{2} A + x^{2} C + 3 x A + x B + 2 x C + 2 A + 2 B + C$

The coefficients near the like terms should be equal, so the following system is obtained:

$\begin{cases} A + C = -4\\3 A + B + 2 C = 13\\2 A + 2 B + C = -12 \end{cases}$

Solving this system, we get that $A=50, B=-29, C=-54$ .

Therefore,

$\frac{- 4 x^{2} + 13 x - 12}{\left(x + 1\right)^{2} \left(x + 2\right)}=\frac{50}{x + 1}+\frac{-29}{\left(x + 1\right)^{2}}+\frac{-54}{x + 2}$

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3 years ago

Find the non permissible replacement for -5/3y

Kamila [148]

Answer:

Step-by-step explanation:

The nonpermissible replacement of the variable in an expression is the value of x that will make the denominator of the expression zero.

This problem will therefore be written as -5/3y≠0, and it can't be 0, so 0 is your non permissible replacement.

Hope this helps!

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3 years ago

Consider the given function.

Misha Larkins [42]

Answer:

x -intercepts are x = -3 and x = 1

y-intercept at y = -3

The line of symmetry is x = -1

vertex is (h,k) = (-1,-4)

Step-by-step explanation:

Given a quadratic equations: h(x) = (x + 1)^2 - 4

The vertex for is :

h(x) = a(x - h)² + k

where,'h' is the axis of symmetry and (h,k) is the vertex.

So from the given equation we will rewrite the equation as:

h(x) = (x - (-1))^2 - 4

Hence,h = -1 and k = -4

The line of symmetry is x = -1 and vertex is (h,k) = (-1,-4)

Now, we have to find the x intercepts,

using the equation,

(x + 1)² = 4

Taking the square root on both sides,

√(x + 1)² = √4

x + 1 = ± 2

x = 1 or x = -3

So x -intercepts are x = -3 and x = 1

For y-intercept put x = 0 into the real equation:

h(x) = (0 +1)^2 - 4

y = 1 - 4

y = -3

So y-intercept at y = -3

Here with I have attached the graph.

Thank you.

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