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2 years ago

PLEASE HELP ILL MAKE YOU THE BRAINIEST What is the value of x

Mathematics

2 answers:

Alex787 [66]2 years ago

8 0

$~~~~~~\cos 45^{\circ} = \dfrac{\text{Base}}{\text{Hypotenuse}}\\ \\\implies \dfrac 1{\sqrt 2} = \dfrac{x}{22}\\\\\implies x = \dfrac{22}{\sqrt 2}\\\\\implies x = \dfrac{22\sqrt 2}{2}\\\\\implies x = 11\sqrt 2\\$

garik1379 [7]2 years ago

7 0

There are two ways to solve this

I’d be happy to answer any questions if you have any.

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100% probability. the probability of an event is 1 (or 100%) if and only if the event can never occur.

Ierofanga [76]

True us the answer too your question.

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3 years ago

Please Help ASAP!! Marking Brainliest!! Divide <img src="https://tex.z-dn.net/?f=12x%5E2%20%2B%204x%5E3%20%2B%2018%20%2B%2016x"

MrMuchimi

Answer:

The answer is C

Step-by-step explanation:

Just calculate it

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2 years ago

I need help with this im currently brainded

makvit [3.9K]

> your welcome :) and tell me if it’s right

4 0

2 years ago

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Drag the tiles to the correct boxes to complete the pairs not all tiles will be used match each quadratic graph to its respectiv

ch4aika [34]

Answer:

Part 1) The function of the First graph is $f(x)=(x-3)(x+1)$

Part 2) The function of the Second graph is $f(x)=-2(x-1)(x+3)$

Part 3) The function of the Third graph is $f(x)=0.5(x-6)(x+2)$

See the attached figure

Step-by-step explanation:

we know that

The quadratic equation in factored form is equal to

$f(x)=a(x-c)(x-d)$

where

a is the leading coefficient

c and d are the roots or zeros of the function

Part 1) First graph

we know that

The solutions or zeros of the first graph are

x=-1 and x=3

The parabola open up, so the leading coefficient a is positive

The function is equal to

$f(x)=a(x-3)(x+1)$

Find the value of the coefficient a

The vertex is equal to the point (1,-4)

substitute and solve for a

$-4=a(1-3)(1+1)$

$-4=a(-2)(2)$

$a=1$

therefore

The function is equal to

$f(x)=(x-3)(x+1)$

Part 2) Second graph

we know that

The solutions or zeros of the first graph are

x=-3 and x=1

The parabola open down, so the leading coefficient a is negative

The function is equal to

$f(x)=a(x-1)(x+3)$

Find the value of the coefficient a

The vertex is equal to the point (-1,8)

substitute and solve for a

$8=a(-1-1)(-1+3)$

$8=a(-2)(2)$

$a=-2$

therefore

The function is equal to

$f(x)=-2(x-1)(x+3)$

Part 3) Third graph

we know that

The solutions or zeros of the first graph are

x=-2 and x=6

The parabola open up, so the leading coefficient a is positive

The function is equal to

$f(x)=a(x-6)(x+2)$

Find the value of the coefficient a

The vertex is equal to the point (2,-8)

substitute and solve for a

$-8=a(2-6)(2+2)$

$-8=a(-4)(4)$

$a=0.5$

therefore

The function is equal to

$f(x)=0.5(x-6)(x+2)$

3 0

3 years ago

Solve the system of linear equations by elimination.<br> -3.x – 5y = -7<br> -4x + 5y = 14

julia-pushkina [17]

Answer:

(- 1, 2 )

Step-by-step explanation:

Given the 2 equations

- 3x - 5y = - 7 → (1)

- 4x + 5y = 14 → (2)

Add the 2 equations term by term to eliminate y , that is

- 7x + 0 = 7

- 7x = 7 ( divide both sides by - 7 )

x = - 1

Substitute x = - 1 into either of the 2 equations and solve for y

Substituting into (1)

- 3(- 1) - 5y = - 7

3 - 5y = - 7 ( subtract 3 from both sides )

- 5y = - 10 ( divide both sides by - 5 )

y = 2

solution is (- 1, 2 )

8 0

2 years ago

Read 2 more answers