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Olenka [21]
3 years ago
8

Help please!!!! Giving out brainliest :((( help me ASAP!!! Thank youu :)

Mathematics
1 answer:
igomit [66]3 years ago
8 0
Help please!!!! Giving out brainliest :((( help me ASAP!!! Thank youu :)

33.51
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Please help me I will give you the brain thing and extra points. (image below) 3/4
Troyanec [42]
First, we have to find the difference between the x and y numbers in the table. By dividing the y by the x, the c is being multiplied by 15.

So the answer would be y=15x

Brainliest please!
4 0
3 years ago
Please help, don’t understand. For #14 what would the input of the function be?
Goryan [66]

a)

h(t) , namely "h" variable is being represented in terms of "t".

h = dependent variable, depends on the value of "t", output.

t = independent variable, takes on values freely, input.


b)

\bf \stackrel{\stackrel{h(t)}{t=1.4}}{h(1.4)} = \stackrel{h=12.8}{12.8}~~ \begin{cases} t=1.4~~&\textit{after 1.4 seconds}\\ h=12.8~~&\textit{the ball is 12.8 meters up} \end{cases}


c)

\bf h(2)=-5(2)^2+14(2)+3\implies h(2)=-20+28+3\implies h(2)=11

after 2 seconds, the ball is 11 meters high up.

5 0
3 years ago
Help how do I expand it​
Free_Kalibri [48]

Answer:

expand what and i will help you have a nice day please give me brainliest:))

Step-by-step explanation:

8 0
3 years ago
Solve the triangles. Round to the nearest tenth А 7 B 6 C​
lisov135 [29]

Answer:

Sure but can you show me the triangle.

7 0
3 years ago
The diagram shows a 5 cm x 5 cm x 5 cm cube.
mylen [45]

Answer:

~8.66cm

Step-by-step explanation:

The length of a diagonal of a rectangular of sides a and b is

\sqrt{a^2+b^2}

in a cube, we can start by computing the diagonal of a rectangular side/wall containing A and then the diagonal of the rectangle formed by that diagonal and the edge leading to A. If the cube has sides a, b and c, we infer that the length is:

\sqrt{\sqrt{a^2+b^2}^2 + c^2} = \sqrt{a^2+b^2+c^2}

Using this reasoning, we can prove that in a n-dimensional space, the length of the longest diagonal of a hypercube of edge lengths a_1, a_2, a_3, \ldots, a_n is

\sqrt{a_1^2 + a_2^2 + a_3^2 + \ldots + a_n^2}

So the solution here is

\sqrt{(5cm)^2 + (5cm)^2 + (5cm)^2} = \sqrt{75cm^2} = 5\sqrt{3cm^2} \approx 5\cdot 1.732cm = 8.66cm

5 0
3 years ago
Read 2 more answers
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