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seraphim [82]
3 years ago
13

Find the value for x​

Mathematics
1 answer:
d1i1m1o1n [39]3 years ago
5 0

Answer:

Value of x --> 15°

<u>Step-by-step explanation:</u>

(2x + 60°) and 6x lie on a straight line. So, they form a linear pair. Their sum gives 180°.

---> (2x + 60°) + 6x = 180°

     2x + 60° + 6x = 180°

     8x + 60° = 180°

     8x = 180°- 60°

     8x = 120

      x = \frac{120}{8}

<h3><u>x equals 15°</u></h3>
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You need at least $535 to go on a trip to California. You have already save $200. You decide to save an additional $25 per week.
FrozenT [24]

Answer:

200+25w

Step-by-step explanation:

200=starting amount

25=every week gaining

w=weeks

Answer= 14.5 weeks

Hope it helps!

4 0
3 years ago
Read 2 more answers
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
Please help if you could
Goshia [24]

Answer:

f(x)= 6x+9....

Step-by-step explanation:

The given equation is:

y-6x-9=0

Add 6x+9 at both sides:

y-6x-9+6x+9=0+6x+9

Solve the like terms:

on the L.H.S -6x will be cancelled out by +6x and -9 will be cancelled out by +9

y=6x+9

Now convert it in function notation:

f(x)=y

f(x)= 6x+9....

3 0
3 years ago
PLEASE HELP!!! I've been really struggling with these. Can someone assist with this?
Archy [21]
When dealing with little odds you must know one part and the whole part. For number 1 a fraction , 7 is a part of a 11 which is the whole. For 2 a percent, 6% part and 100% whole, here you had to know that for 6% to exist there had to be a 100% . For 3 a ratio, the first number is part (2) and the second is whole (7). Look at pic for all work.

4 0
3 years ago
The perimeter of a semicircle is 20.56 yards. What is the semicircle's radius?
Scorpion4ik [409]

Answer:

Step-by-step explanation:

Perimeter = 20.56mm

Given the perimeter, find the diameter

= 20.56

D = 20.56 ÷

D = 6.54 mm

Perimeter of the Semicircle = half the circle arc + the straight line (Diameter)

Semicircle = 0.5(20.56) + 6.54cm

Semicircle = 16.83mm

7 0
3 years ago
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