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tigry1 [53]
3 years ago
11

11+8q=3q-19 I need help please idk the answer

Mathematics
1 answer:
iren2701 [21]3 years ago
4 0

Answer:

-6

Step-by-step explanation:

1) combine all like terms

(bring 3q to the left and 11 to the right so you can combine the like terms)

(doing that will make 3q negative and 11 negative since your moving them so their sign changes as well)

8q-3q=-19-11

5q=-30

Q=-6

Hope this helps

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For which system of equations is (2, 2) a solution?
Eva8 [605]
Answer D.

Insert (2, 2) in for x and y in the equations.

2(2)+ 3(2) = ?
? = 10
4(2) + 5(2) = ?
? = 18
8 0
3 years ago
A plane descends 1000 ft while flying<br> 2.7 mi. What is the angle of descent?
Murrr4er [49]

Answer:

4.02°

Step-by-step explanation:

The angle of descent can be obtained using trigonometry ;

Since we have a right angled triangle ;

Let's use the relation:

Sin θ = opposite / hypotenus

Hypotenus = 2.7 miles

1 mile = 5280 feets

2.7 miles = (2.7 * 5280) = 14256 feets

Sin θ = 1000 / 14256

Sin θ = 0.0701459

θ = sin^-1(0.0701459)

θ = 4.0223674

3 0
3 years ago
Baytown Village Stone Creations is making a custom stone bench. the recommended height for the bench is 18 in. the depth of the
Natalija [7]

Answer:

x = tex]5\frac{\textup{7}}{\textup{8}}\ in[/tex]

Step-by-step explanation:

Given:

Total height of the bench = 18 in

Depth of the stone bench = 3\frac{\textup{3}}{\textup{8}}\ in = \frac{\textup{27}}{\textup{8}}\ in  = 3.375 in

Measure of stones = 3\frac{\textup{1}}{\textup{2}}\ in = \frac{\textup{7}}{\textup{2}}\ in  = 3.5 in

measure of another stone =  5\frac{\textup{1}}{\textup{4}}\ in = \frac{\textup{21}}{\textup{4}}\ in  = 5.25 in

let the height of the third stone be 'x'

Now,

The total height of the bench = depth of bench + Measure of two stones + x

18 = 3.375 + 3.5 + 5.25 + x

or

x = 18 - 12.125

or

x = 5.875 in

or

x = tex]5\frac{\textup{7}}{\textup{8}}\ in[/tex]

6 0
3 years ago
Please help!!<br> Write a matrix representing the system of equations
frozen [14]

Answer:

(4, -1, 3)

Step-by-step explanation:

We have the system of equations:

\left\{        \begin{array}{ll}            x+2y+z =5 \\    2x-y+2z=15\\3x+y-z=8        \end{array}    \right.

We can convert this to a matrix. In order to convert a triple system of equations to matrix, we can use the following format:

\begin{bmatrix}x_1& y_1& z_1&c_1\\x_2 & y_2 & z_2&c_2\\x_3&y_2&z_3&c_3 \end{bmatrix}

Importantly, make sure the coefficients of each variable align vertically, and that each equation aligns horizontally.

In order to solve this matrix and the system, we will have to convert this to the reduced row-echelon form, namely:

\begin{bmatrix}1 & 0& 0&x\\0 & 1 & 0&y\\0&0&1&z \end{bmatrix}

Where the (x, y, z) is our solution set.

Reducing:

With our system, we will have the following matrix:

\begin{bmatrix}1 & 2& 1&5\\2 & -1 & 2&15\\3&1&-1&8 \end{bmatrix}

What we should begin by doing is too see how we can change each row to the reduced-form.

Notice that R₁ and R₂ are rather similar. In fact, we can cancel out the 1s in R₂. To do so, we can add R₂ to -2(R₁). This gives us:

\begin{bmatrix}1 & 2& 1&5\\2+(-2) & -1+(-4) & 2+(-2)&15+(-10) \\3&1&-1&8 \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 2& 1&5\\0 & -5 & 0&5 \\3&1&-1&8 \end{bmatrix}

Now, we can multiply R₂ by -1/5. This yields:

\begin{bmatrix}1 & 2& 1&5\\ -\frac{1}{5}(0) & -\frac{1}{5}(-5) & -\frac{1}{5}(0)& -\frac{1}{5}(5) \\3&1&-1&8 \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\3&1&-1&8 \end{bmatrix}

From here, we can eliminate the 3 in R₃ by adding it to -3(R₁). This yields:

\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\3+(-3)&1+(-6)&-1+(-3)&8+(-15) \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\0&-5&-4&-7 \end{bmatrix}

We can eliminate the -5 in R₃ by adding 5(R₂). This yields:

\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\0+(0)&-5+(5)&-4+(0)&-7+(-5) \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\0&0&-4&-12 \end{bmatrix}

We can now reduce R₃ by multiply it by -1/4:

\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\ -\frac{1}{4}(0)&-\frac{1}{4}(0)&-\frac{1}{4}(-4)&-\frac{1}{4}(-12) \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\0&0&1&3 \end{bmatrix}

Finally, we just have to reduce R₁. Let's eliminate the 2 first. We can do that by adding -2(R₂). So:

\begin{bmatrix}1+(0) & 2+(-2)& 1+(0)&5+(-(-2))\\ 0 & 1 & 0& -1 \\0&0&1&3 \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 0& 1&7\\ 0 & 1 & 0& -1 \\0&0&1&3 \end{bmatrix}

And finally, we can eliminate the second 1 by adding -(R₃):

\begin{bmatrix}1 +(0)& 0+(0)& 1+(-1)&7+(-3)\\ 0 & 1 & 0& -1 \\0&0&1&3 \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 0& 0&4\\ 0 & 1 & 0& -1 \\0&0&1&3 \end{bmatrix}

Therefore, our solution set is (4, -1, 3)

And we're done!

3 0
3 years ago
Si el largo de un rectángulo mide el doble del ancho que es "a", ¿cuál es su área?
rosijanka [135]

Answer:

El área es 2 veces el cuadrado del ancho del rectángulo.

Step-by-step explanation:

El área de un rectángulo viene dado por:

A = a*l

En donde:

a: es el ancho

l: es el largo

Si el largo es el doble del ancho:

l = 2a

Entonces el área es:

A = a*l = a*(2a) = 2a^{2}

Por lo tanto, el área es 2 veces el cuadrado del ancho del rectángulo.

Espero que te sea de utilidad!  

6 0
3 years ago
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