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3 years ago

Solve for x in the equation 2x^2-4x-9=29

Mathematics

1 answer:

Andrews [41]3 years ago

3 0

Answer:

x = 1 ±2sqrt(5)

Step-by-step explanation:

2x^2-4x-9=29

Add 9 to each each side

2x^2-4x-9+9=29+9

2x^2-4x=38

Divide by 2

2/2x^2-4/2x=38/8

x^2 -2x =19

Complete the square

x^2 -2x + (-2/2)^2 = 19 +(-2/2)^2

x^2 -2x +1 = 19+1

(x-1)^1=2 = 20

Take the square root of each side

sqrt((x-1)^2) = ±sqrt(20)

x-1 = ±sqrt(20)

Add 1 to each side

x-1+1 = 1 ±sqrt(20)

x = 1 ±sqrt(20)

Simplifying the square root of 20

x = 1 ±sqrt(4)sqrt(5)

x = 1 ±2sqrt(5)

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WILL MARK YOU BRAINLIEST!!!!!!

My name is Ann [436]

Answer:

y = 88

Step-by-step explanation:

Don't worry about the other stuff, just worry about what's in the triangle that has Y.

53°, 39°, and y°

the sum should be equal to 180 so add 53 adn 39 and subtract it from 180.

180 - (53 + 39) = y

180 - 92 = y

88 = y

Hope this helped! Pls Brainliest my answer! It would help amd mean a lot! ;)

6 0

3 years ago

Find the length of FT¯¯¯¯¯¯¯ A. 77.71 B. 72.47 C. 56.84 D. 49.42

Mkey [24]

Answer:

D, 49.42

Step-by-step explanation:

ΔVFT=180-90-43=47

formula

a/sin A = b/sin B/ = c/sin C

So,

FV/sin90=53/sin47

FV=72.4684

FT=√(72.4684)^2-(53)^2

FT=49.4234

Ans:D

5 0

3 years ago

Read 2 more answers

Use the graph below to estimate the solution to the system of equations shown.

Mila [183]

Answer: The second choice is answer.

Step-by-step explanation:

<h3>The first line (Blue One)</h3>

$y=-x-3$ (Slope is -1 and intercepts y at -3)

<h3>The second one (Red One)</h3>

$y=2x-8$ (Slope is 2, to find y-intercept. Substitute x = 0 then we get -8) (For x-intercept, it's 4. Just substitute y=0 to get x-intercept.)

Because there are choices, it's easy to notice.

The first one, x = -4 and almost -5 which is not right. (x is around 1 almost 2.)

The second one, x = 1 and almost 2 is right.

The third one, x = -1 and almost -2 is not right (The intersection isn't even at -1 or any negative number for x-axis.)

The fourth one, just like the first one... x = 4 almost 5

<h3>Or solve the equations.</h3>

$y=-x-3\\y=2x-8$

Substitute y=2x-8 in y=-x-3

$2x-8=-x-3\\2x-8+x+3=0\\3x-5=0\\3x=5\\x=\frac{5}{3}\\(x=1\frac{2}{3})$

Look at the value of x then find the answer that matches the value of x.

Now for y, substitute x = 5/3 in y = -x-3 (or in 2x-8 if you want.)

$y=-\frac{5}{3} -3\\y=-\frac{5}{3} -\frac{9}{3} \\y=-\frac{14}{3} \\(y=-4\frac{2}{3} )$

So the answer is the second choice.

6 0

3 years ago

A surveyor leaves her base camp and drives 42km on a bearing of 032degree she then drives 28km on a bearing of 154degree,how far

ValentinkaMS [17]

Answer:

The surveyor is 36.076 kilometers far from her camp and her bearing is 16.840° (standard form).

Step-by-step explanation:

The final position of the surveyor is represented by the following vectorial sum:

$\vec r = \vec r_{1} + \vec r_{2} + \vec r_{3}$ (1)

And this formula is expanded by definition of vectors in rectangular and polar form:

$(x,y) = r_{1}\cdot (\cos \theta_{1}, \sin \theta_{1}) + r_{2}\cdot (\cos \theta_{2}, \sin \theta_{2})$ (1b)

Where:

$x, y$ - Resulting coordinates of the final position of the surveyor with respect to origin, in kilometers.

$r_{1}, r_{2}$ - Length of each vector, in kilometers.

$\theta_{1}, \theta_{2}$ - Bearing of each vector in standard position, in sexagesimal degrees.

If we know that $r_{1} = 42\,km$ , $r_{2} = 28\,km$ , $\theta_{1} = 32^{\circ}$ and $\theta_{2} = 154^{\circ}$ , then the resulting coordinates of the final position of the surveyor is:

$(x,y) = (42\,km)\cdot (\cos 32^{\circ}, \sin 32^{\circ}) + (28\,km)\cdot (\cos 154^{\circ}, \sin 154^{\circ})$

$(x,y) = (35.618, 22.257) + (-25.166, 12.274)\,[km]$

$(x,y) = (10.452, 34.531)\,[km]$

According to this, the resulting vector is locating in the first quadrant. The bearing of the vector is determined by the following definition:

$\theta = \tan^{-1} \frac{10.452\,km}{34.531\,km}$

$\theta \approx 16.840^{\circ}$

And the distance from the camp is calculated by the Pythagorean Theorem:

$r = \sqrt{(10.452\,km)^{2}+(34.531\,km)^{2}}$

$r = 36.078\,km$

The surveyor is 36.076 kilometers far from her camp and her bearing is 16.840° (standard form).

5 0

3 years ago

Which method do you prefer to use to find sums- count by tens and ones, use compatible numbers, or use friendly numbers and adju

Usimov [2.4K]

I prefer to use compatible numbers because by using this method it is easier to make a sum mentally. This is true because compatible numbers are close in value to the actual numbers. For a better understanding, let's take an example:

Suppose you have two numbers, namely 640 and 40. These two numbers are compatible for division because:

64 ÷ 4 = 16

So, we have used mental arithmetic to solve a more complex problem.

7 0

3 years ago

Read 2 more answers