Please answer 23 and 24 for me pls pls pls

Write a fraction equivalent to 60mi/4hr that has 2 in the denominator

lara [203]

30mi/2hr

Explanation: just divide both by two. It won’t always work like this. You can’t always divide something by 2 to get 2. Find a number that will go into the denominator two times and divide the numerator by it also.

5 0

4 years ago

In a right angle triangle the side lengths are 4 and 2. Please find the hypotenuse. You

anygoal [31]

Answer:

$\sqrt{20}$ or 4.47

Step-by-step explanation:

$a^{2}+^{2}=c^{2} \\4^{2}+2^{2}=c^{2}\\16+4=c^{2}\\c=\sqrt{20}$

4 0

3 years ago

What is the GCF of two prime numbers?

ehidna [41]

Prime numbers: numbers which only have the factors of themselves and 1.

They include numbers such as 17, 19, 23, etc. These are numbers which only have two factors: such as 17, with the factors only being 17 and 1.

So therefore, this means, that for any two prime numbers, unless they are the same number, the only factor they will share is 1. So therefore the GCF of two prime numbers is 1.

Hope this helped!

4 0

4 years ago

the verticales of triangle XYZ are X(-3,-6),Y(21,-6),and Z(21,4). What is the perimeter of the triangle

andrezito [222]

Answer:

Step-by-step explanation:

The formula of a distance between two points:

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

X(-3, -6), Y(21, -6). Substitute:

$XY=\sqrt{21-(-3))^2+(-6-(-6))^2}=\sqrt{24^2+0^2}=\sqrt{24^2}=24$

X(-3, -6), Z(21, 4). Substitute:

$XZ=\sqrt{(21-(-3))^2+(4-(-6))^2}=\sqrt{24^2+10^2}=\sqrt{576+100}=\sqrt{676}=26$

Y(21, -6), Z(21, 4). Substitute:

$YZ=\sqrt{(21-21)^2+(4-(-6))^2}=\sqrt{0^2+10^2}=\sqrt{10^2}=10$

The perimeter of the triangle XYZ:

$P_{\triangle XYZ}=XY+XZ+YZ$

Substitute:

$P_{\triangle XYZ}=24+26+10=60$

6 0

3 years ago

If vector u has its initial point at (-7, 3) and its terminal point at (5, -6), u =

attashe74 [19]

First of all, let θθ be some angle in (0,π)(0,π). Then

θθ is acute ⟺⟺ θ<π2θ<π2 ⟺⟺ cosθ>0cos⁡θ>0.θθ is right ⟺⟺ θ=π2θ=π2 ⟺⟺ cosθ=0cos⁡θ=0.θθ is obtuse ⟺⟺ θ>π2θ>π2 ⟺⟺ cosθ<0cos⁡θ<0.

Now, to see if (say) angle AA of the triangle ABCABC is acute/right/obtuse, we need to check whether cos∠BACcos⁡∠BAC is positive/zero/negative. But what is cos∠BACcos⁡∠BAC? It is the angle made by the vectors AB−→−AB→ and AC−→−AC→. (When you are computing the angle at a particular vertex vv, you should make sure that both the vectors corresponding to the two adjacent sides have that vertex vv as the initial point.) We will first compute these two vectors:

AB−→−=(0,0,0)−(1,2,0)=(−1,−2,0)AB→=(0,0,0)−(1,2,0)=(−1,−2,0)AC−→−=(−2,1,0)−(1,2,0)=(−3,−1,0)AC→=(−2,1,0)−(1,2,0)=(−3,−1,0)Therefore, the angle between these vectors is given by:cos∠BAC=AB−→−⋅AC−→−|AB−→−||AC−→−|=…(1)(1)cos⁡∠BAC=AB→⋅AC→|AB→||AC→|=…Can you take it from here? From the sign of this value, you should be able to decide if angle AA is acute/right/obtuse.

Now, do the same procedure for the remaining two angles BB and CC as well. That should help you solve the problem.

A shortcut. Since you are not interested in the actual values of the angles, but you need only whether they are acute, obtuse or right, it is enough to compute only the sign of the numerator (the dot product between the vectors) in formula (1). The denominator is always positive.

6 0

4 years ago