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3 years ago

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First answer that is right gets brainliest. What is the value of x in the equation 4(2x + 10) = 0? (1 point) Group of answer cho

ices −5 −7 5 7

1 answer:

Makovka662 [10]3 years ago

5 0

The value of x in given equation 4(2x + 10) = 0 is -5

<em><u>Solution:</u></em>

Given that we have to find the value of "x" in the given equation

Given equation is:

4(2x + 10) = 0

The value of the variable is found by adding, subtracting, multiplying or dividing both sides of the equation to simplify the equation and isolate the variable.

The goal is to have the variable on one side of the equation and numbers on the other.

Let us first multiply 4 with terms inside bracket

4(2x + 10) = 0

8x + 40 = 0

8x = -40

$x = \frac{-40}{8} = -5$

Thus the value of "x" is -5

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What is the length of AC

meriva

Hi!
This is a fun one, as it delves into basic trigonometry.

We're going to use the Pythagorean theorem here, which says that for right triangles where "c" is the hypotenuse,

a² + b² = c²

We have to split this large triangle into two parts, both of which are right triangles. (This is why they drew a line in the middle to tell you that the larger triangle is composed of two right triangles.)

Let's do the one on the right first.

We know that the length of the hypotenuse is 10, and that the length of one of the legs is 6.5. If we plug this into our equation, we'll get the length of the other leg. I'm choosing "b" to be 6.5, but it really doesn't matter if you pick "a" or "b", so long as you reserve "c" for the hypotenuse (longest side).

a² + 6.5² = 10²
a² + 42.25 = 100
a² = 57.75
√a² = √57.75
a ≈ 7.6

Therefore, the length of DC is about 7.6.

Find the length of AD using the same method (7.5 is the hypotenuse "c", and 6.5 is one of the legs "a" or "b"). Then, once you have AD, add the lengths of AD and DC to get AC.

Have a great one!

7 0

3 years ago

What is the volume of the following triangular prism?<br> 60 cm3<br> 12 cm3<br> 3.75 cm3<br> 6 cm3

DochEvi [55]

By definition, the prism volume is given by:
$V = Ab * L
$ Where,
Ab: base area
L: long
Substituting values we have:
$V = ((1/2) * (12) * (2.5)) * (4)
 
 V = 60 cm ^ 3$
Answer:
The volume of the triangular prism is:
$V = 60 cm ^ 3$

5 0

3 years ago

Identify this conic section. 16y = x^2

kiruha [24]

ANSWER

A parabola.

EXPLANATION

The given conic is :

$16y = {x}^{2}$

This can be rewritten as:

${x}^{2} = 16y$

${x}^{2} = 4(4)y$

This is a parabola with the vertex at the origin.

The foci is (0,4)

Therefore the given conic section is a parabola that has an axis of symmetry parallel to the y-axis.

7 0

3 years ago

1) For question 1, use the figure below. a) Are the following polygons similar? Explain. (2 pts) Yes or No b) Assume the polygon

Delvig [45]

Answer: a) Yes b) Polygon ABCD ~ Polygon EFGH c) JKLM is 3x bigger than polygon ABCD

Step-by-step explanation:

(for part c)

JM= 36

AD= 12

36/12= 3

JKLM is 3x bigger than polygon ABCD

3 0

3 years ago

Solve y ' ' + 4 y = 0 , y ( 0 ) = 2 , y ' ( 0 ) = 2 The resulting oscillation will have Amplitude: Period: If your solution is A

Vlad [161]

Answer:

$y(x)=sin(2x)+2cos(2x)$

Step-by-step explanation:

$y''+4y=0$

This is a homogeneous linear equation. So, assume a solution will be proportional to:

$e^{\lambda x} \\\\for\hspace{3}some\hspace{3}constant\hspace{3}\lambda$

Now, substitute $y(x)=e^{\lambda x}$ into the differential equation:

$\frac{d^2}{dx^2} (e^{\lambda x} ) +4e^{\lambda x} =0$

Using the characteristic equation:

$\lambda ^2 e^{\lambda x} + 4e^{\lambda x} =0$

Factor out $e^{\lambda x}$

$e^{\lambda x}(\lambda ^2 +4) =0$

Where:

$e^{\lambda x} \neq 0\\\\for\hspace{3}any\hspace{3}\lambda$

Therefore the zeros must come from the polynomial:

$\lambda^2+4 =0$

Solving for $\lambda$ :

$\lambda =\pm2i$

These roots give the next solutions:

$y_1(x)=c_1 e^{2ix} \\\\and\\\\y_2(x)=c_2 e^{-2ix}$

Where $c_1$ and $c_2$ are arbitrary constants. Now, the general solution is the sum of the previous solutions:

$y(x)=c_1 e^{2ix} +c_2 e^{-2ix}$

Using Euler's identity:

$e^{\alpha +i\beta} =e^{\alpha} cos(\beta)+ie^{\alpha} sin(\beta)$

$y(x)=c_1 (cos(2x)+isin(2x))+c_2(cos(2x)-isin(2x))\\\\Regroup\\\\y(x)=(c_1+c_2)cos(2x) +i(c_1-c_2)sin(2x)\\$

Redefine:

$i(c_1-c_2)=c_1\\\\c_1+c_2=c_2$

Since these are arbitrary constants

$y(x)=c_1sin(2x)+c_2cos(2x)$

Now, let's find its derivative in order to find $c_1$ and $c_2$

$y'(x)=2c_1 cos(2x)-2c_2sin(2x)$

Evaluating $y(0)=2$ :

$y(0)=2=c_1sin(0)+c_2cos(0)\\\\2=c_2$

Evaluating $y'(0)=2$ :

$y'(0)=2=2c_1cos(0)-2c_2sin(0)\\\\2=2c_1\\\\c_1=1$

Finally, the solution is given by:

$y(x)=sin(2x)+2cos(2x)$

5 0

4 years ago