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DedPeter [7]
3 years ago
5

Super confused, appreciate it if someone could help lol​

Mathematics
2 answers:
Leona [35]3 years ago
6 0

Answer: the 1 2 and 3

Step-by-step explanation:

zhenek [66]3 years ago
6 0

Answer:

bruv edginuity sucks haha

Step-by-step explanation:

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U<br>12 - 7 + 2<br>15 15<br>15<br>1​
Dennis_Churaev [7]

Answer:

I didn't know

6 0
2 years ago
If aa skateboard ramp is 15 meters long and it extends 12 meters from the base of the starting point. How high is the ramp?
guapka [62]

The height of the ramp is 9 meter

<u>Solution:</u>

It is given that a skateboard ramp is 15 meters long and it extends 12 meters from the base of the starting point.

If we look at the sum, closely we understand that the ramp is a right angled triangle.

Which has a base length of 12 metres and a hypotenuse of 15 metres.

We need to find its height.

To do so we can use the Pythagoras theorem

Pythagorean theorem, states that the square of the length of the hypotenuse is equal to the sum of squares of the lengths of other two sides of the right-angled triangle.

By above definition, we get

\text {Hypotenuse}^{2}=\text {height}^{2}+\text { base }^{2}

Since we already know the lengths of the hypotenuse and base we can substitute them in the formula and solve for the height.

Let height of the ramp be denoted by ‘h’

\begin{array}{l}{15^{2}=12^{2}+h^{2}} \\\\ {225=144+h^{2}} \\\\ {225-144=h^{2}} \\\\ {81=h^{2}} \\\\ {h=\sqrt{81}} \\\\ {h=9}\end{array}

Therefore, the height of the ramp is 9 meter

3 0
3 years ago
4. How fast was a driver going if the car left skid marks that were 32 feet long on dry concrete? (The coefficient of friction i
Monica [59]

Answer:

v_{o} \approx 31.247\,mph

Step-by-step explanation:

The deceleration experimented by the car is the product of kinetic coefficient of friction and gravitational constant:

a = \mu_{k}\cdot g

a = (1.02)\cdot \left(32.174\,\frac{ft}{s^{2}} \right)

a = 32.817\,\frac{ft}{s^{2}}

The initial speed is computed from the following kinematic expression:

v^{2} = v_{o}^{2} - 2\cdot a \cdot \Delta s

v_{o} = \sqrt{v^{2}+2\cdot a\cdot \Delta s}

v_{o} = \sqrt{\left(0\,\frac{ft}{s} \right)^{2}+2\cdot \left(32.817\,\frac{ft}{s^{2}} \right)\cdot (32\,ft)}

v_{o}\approx 45.829\,\frac{ft}{s}

v_{o} \approx 31.247\,mph

8 0
3 years ago
Solve for the solution: 2m+1≥7
QveST [7]

Answer:

m≥3

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
Find the degree of the polynomial: a^3+3a^2−5a
emmainna [20.7K]

Answer:

The degree of the polynomial is 3

Step-by-step explanation:

Given:

a^3+3a^2-5a

To Find:

The degree of the polynomial= ?

Solution:

The degree of the polynomial   is the value of the greatest exponent of any expression (except the constant ) in the polynomial. To find the degree all that you have to do is find the largest exponent in the polynomial

Here in the given polynomial

a^3+3a^2- 5a

The terms are

a^3

3a^2

5a

The term  a^3 has  the largest exponent of  3

Note:  The degree of the polynomial does not depend  on coefficients of the terms

3 0
3 years ago
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