Ask question

Ask question

All categories

3 years ago

11

Solve for x ~

"x {}^{2} - 5x + 6 = 0" alt="x {}^{2} - 5x + 6 = 0" align="absmiddle" class="latex-formula">

ty! :)

1 answer:

alexdok [17]3 years ago

4 0

<h2>ANSWER!</h2>

Factor x² - 5x +6

<h3>(x−3)(x−2)=0</h3>

Solve for x

<h2>x=3,2</h2>

therefore the answer is x=3,2

You might be interested in

I am Lyosha [343]

Option C

The ratio for the volumes of two similar cylinders is 8 : 27

<h3><u>Solution:</u></h3>

Let there are two cylinder of heights "h" and "H"

Also radius to be "r" and "R"

$\text { Volume of a cylinder }=\pi r^{2} h$

Where π = 3.14 , r is the radius and h is the height

Now the ratio of their heights and radii is 2:3 .i.e

$\frac{\mathrm{r}}{R}=\frac{\mathrm{h}}{H}=\frac{2}{3}$

<em><u>Ratio for the volumes of two cylinders</u></em>

$\frac{\text {Volume of cylinder } 1}{\text {Volume of cylinder } 2}=\frac{\pi r^{2} h}{\pi R^{2} H}$

Cancelling the common terms, we get

$\frac{\text {Volume of cylinder } 1}{\text {Volume of cylinder } 2}=\left(\frac{\mathrm{r}}{R}\right)^{2} \times\left(\frac{\mathrm{h}}{\mathrm{H}}\right)$

Substituting we get,

$\frac{\text {Volume of cylinder } 1}{\text {Volume of cylinder } 2}=\left(\frac{2}{3}\right)^{2} \times\left(\frac{2}{3}\right)$

$\frac{\text {Volume of cylinder } 1}{\text {Volume of cylinder } 2}=\frac{2 \times 2 \times 2}{3 \times 3 \times 3}$

$\frac{\text {Volume of cylinder } 1}{\text {Volume of cylinder } 2}=\frac{8}{27}$

Hence, the ratio of volume of two cylinders is 8 : 27

7 0

3 years ago

sasho [114]

Answer:

3 a 20 - class yoga package for 260$

3 0

3 years ago

Which expression is equivalent to (x Superscript one-half Baseline y Superscript negative one-fourth Baseline z) Superscript neg

Darina [25.2K]

By using exponent properties, we will get the simplified expression:

$x^{-1}*y^{1/2}*z^2$

<h3>How to simplify the given expression?</h3>

Here we have the expression:

$(x^{1/2}*y^{-1/4}*z)^{-2}$

Remember the exponent properties:

$(a^n)^m = a^{n*m}$

And:

$(a*b)^n = (a^n)*(b^n)$

So using these two properties, we can rewrite:

$(x^{1/2})^{-2}*(y^{-1/4})^{-2}*(z)^{-2}\\\\(x^{-2*1/2})*(y^{-2*-1/4})*(z^{-2}})\\\\x^{-1}*y^{1/2}*z^2$

So we conclude that the completely simplified expression is:

$x^{-1}*y^{1/2}*z^2$

If you want to learn more about exponents:

brainly.com/question/8952483

#SPJ1

8 0

2 years ago

I need help ASAP pls help me!!:/

Olin [163]

Answer:

if i am right it yes yes no

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

Need help ASAP <br> questions in the pics i have sent

Cloud [144]

Answer:

Step-by-step explanation:

Picture 1

In right triangle ABC,

Side AB is the opposite side of angle C.

Picture 2

In triangle MKL,

tan(∠M) = $\frac{\text{Opposite side}}{\text{Adjacent side}}$

= $\frac{KL}{KM}$

= $\frac{15}{8}$

Option (1) is the answer.

Picture 3

In ΔXYZ,

sin(∠Z) = $\frac{\text{Opposite side}}{\text{Hypotenuse}}$

= $\frac{XY}{XZ}$

For the length of XY we will apply Pythagoras theorem in ΔXYZ,

XZ² = XY² + YZ²

XY² = XZ² - YZ²

= (40)² - (32)²

XY = √576

= 24

sin(Z) = $\frac{24}{40}$

sin(Z) = $\frac{3}{5}$

Picture 4

In right triangle DEF,

Cos(D) = $\frac{\text{Adjacent side}}{\text{Hypotenuse}}$

= $\frac{EF}{DF}$

= $\frac{75}{72}$

= $\frac{25}{24}$

Picture 5

In ΔABC,

tan(63°) = $\frac{\text{Opposite side}}{\text{Adjacent side}}$

tan(63°) = $\frac{BC}{AB}$

AB = $\frac{BC}{\text{tan}(63)}$

AB = $\frac{8}{\text{tan}(63)}$

AB = 4.0762 ≈ 4 m

Option (3) will be the answer.

7 0

3 years ago