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Greeley [361]
3 years ago
14

Find the GCF for the given numbers 10 and 12

Mathematics
1 answer:
gogolik [260]3 years ago
6 0

Answer:

factors of 10 are 2

factors of 12 are 2

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Can someone help me with this asap, please and thank you (pic attached) *math*
Jet001 [13]

Step-by-step explanation:

the "x-2" makes it shift by 2 to the right (functional values are created at x that "should have happened" at 2 x units before, so the whole graph moves to the right).

the 0.4 factor compresses it by the factor 0.4

the "-" at the beginning just "flips" (reflects) it to the other side of the x-axis.

3 0
2 years ago
the hypotenuse of a right triangle is 7 cm and the one of leg is 13 inches. find the length of the other leg
Ann [662]

Assuming x represents the length of the first side.

Use Pythagorean theorem to solve:  a^2 + b^2 = c^2

Solve the quadratic equation using the quadratic formula.

At the end, you will get side lengths of #(-14 ± 34) / 4, or -12 and 5

SInce a negative triangle length is impossible, 5 is the value of x and 5 + 7 is the value of x + 7, which makes 12.

The formula for area of a right triangle is A = B H/2

After doing your work you should get. 30cm^2


3 0
3 years ago
Consider the following planes. x + y + z = 6, x + 7y + 7z = 6 (a) Find parametric equations for the line of intersection of the
Lelechka [254]

Answer:

a.x=6,y=-6t,z=6t

b.\theta=29.5^{\circ}

Step-by-step explanation:

We are given that

x+y+z=6

x+7y+7z=6

a.Substitute z=0

x+y=6...(1)

x+7y=6..(2)

Subtract equation (1) from equation (2)

6y=0

y=0

Substitute y=0 in equation(1)

x=6

The point (6,0,0) lie on a line.

r_0=(x_0,y_0,z_0)=(6,0,0)

Let A=

B=

A\times B=\begin{vmatrix}i&j&k\\1&1&1\\1&7&7\end{vmatrix}

A\times B=i(7-7)-j(7-1)+k(7-1)=-6j+6k

Therefore, the vector a'=(a,b,c)=

Line is parallel to vector a' and passing through the point (6,0,0).

The parametric equation is given by

x=x_0+at,y=y_0+bt,z=z_0+ct

Using the formula

The parametric equation is given by

x=6,y=-6t,z=6t

Angle between two plane

a_1x+b_1y+c_1z=d_1

and a_2x+b_2y+c_2z=d_2

cos\theta=\frac{(a_1,b_1,c_1)\cdot (a_2,b_2,c_2)}{\sqrt{a^2_1+b^2_1+c^2_1}\cdot \sqrt{a^2_2+b^2_2+c^2_2}}

Using the formula

cos\theta=\frac{(1,1,1)\cdot(1,7,7)}{\sqrt{1+1+1}\times \sqrt{1+7^2+7^2}}

cos\theta=\frac{1+7+7}{\sqrt 3\times 3\sqrt{11}}

cos\theta=\frac{15}{3\sqrt{33}}}=\frac{5}{\sqrt{33}}

\theta=cos^{-1}(0.87)=29.5^{\circ}

Where \theta in degree.

3 0
3 years ago
What percentage is shown in the picture?
levacccp [35]
Just do 10×8 and you'll get the answer and the green is 24 If you count
8 0
3 years ago
Which inequality is graphed on the coordinate plane?
ruslelena [56]

Answer:

A: y < -2x - 1

Step-by-step explanation:

First, to figure out which part of the graph to be looking at, you will look at the shaded region as the solution area.

So, the solution area is to the left of the line. Said line is a dotted line which means that the points on the line can not be a solution.

If the points on the line can not be a solution then it eliminates answer choices C and D because y can not equal the other side of the inequality.

Now, we will bring the shaded region back into focus to find out if y is greater than or less than the other side. The shaded area is "below" the line so it is less than -2x - 1.

To ensure this is correct, plug two points into the inequality we just chose, one from the shaded region and one not, just to make sure the answer is correct.

I picked (-3,4) to test. So plug them into the inequality to get:

4 < -2 (-3) - 1

So 4 < 6 - 1 which is correct.

Then from the other side of the line I picked (5,2)

2 < -2 (5) - 1

2 < -10 - 1

2 < -11 is not true so the inequality that we found was correct.

8 0
3 years ago
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