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3 years ago

Find the missing side length. 12 15

Mathematics

2 answers:

sashaice [31]3 years ago

7 0

Answer:

you need to show the shape or somet or else I can't rlly help

Leona [35]3 years ago

3 0

Answer:

If the hypotenuse is 15 and one of the sides is 12, the answer is 9.

If these are the side lengths, the hypotenuse is square root of 369, which is about 19.21

Step-by-step explanation:

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Arrange the numbers in the ascending order: 3/4,1/12,9/16,7/24

OLEGan [10]

Ascending means going up
First change into decimals
3/4=0.75
1/12=0.083
9/16=0.5625
7/24=0.291666…..

Start from smallest to largest number
0.291666…..,0.5625,0.75,0.083

Find their fractions
7/24,9/16,3/4,1/12

Answer:7/24,9/16,3/4,1/12

4 0

1 year ago

A bag contains 8 pieces of paper, each with one of these letters written on it: A, B, C, E, K, L, M, and R. One piece of

Delvig [45]

Answer:

Below.

Step-by-step explanation:

Letter P is drawn - Impossible because there is no paper with P on it.

Letter drawn is a consonant - Likely because there are 6 out of 8 letters which are consonants.

Letter drawn is a vowel - Unlikely because there are only 2 vowels out of 8 letters.

Letter drawn is one of the first ten letters in the alphabet - Equally likely as there is 4 of these out of a total of 8.

Letter is L - Unlikely as there is only 1 piece of paper with L on it.

The paper draw has one letter on it - Certain as there is a letter on all the pieces of paper.

6 0

4 years ago

Read 2 more answers

What is the answer please.... need help

mina [271]

Answer:

$f\left(x\right)=x^3-6x^2+3x+10$ is the function of the least degree has the real coefficients and the leading coefficients of 1 and with the zeros -1, 5, and 2.

Step-by-step explanation:

Given the function

$f\left(x\right)=x^3-6x^2+3x+10$

As the highest power of the x-variable is 3 with the leading coefficients of 1.

So, it is clear that the polynomial function of the least degree has the real coefficients and the leading coefficients of 1.

solving to get the zeros

$f\left(x\right)=x^3-6x^2+3x+10$

$0=x^3-6x^2+3x+10$ ∵ $f(x)=0$

$Factor\:x^3-6x^2+3x+10\::\:\left(x+1\right)\left(x-2\right)\left(x-5\right)=0$

$\left(x+1\right)\left(x-2\right)\left(x-5\right)=0$

Using the zero factor principle

if $ab=0\:\mathrm{then}\:a=0\:\mathrm{or}\:b=0\:\left(\mathrm{or\:both}\:a=0\:\mathrm{and}\:b=0\right)$

$x+1=0\quad \mathrm{or}\quad \:x-2=0\quad \mathrm{or}\quad \:x-5=0$

$x=-1,\:x=2,\:x=5$

Therefore, the zeros of the function are:

$x=-1,\:x=2,\:x=5$

$f\left(x\right)=x^3-6x^2+3x+10$ is the function of the least degree has the real coefficients and the leading coefficients of 1 and with the zeros -1, 5, and 2.

Therefore, the last option is true.

8 0

3 years ago

Write the equation of the line that passes through the points (-6,-1) and (-4,2). Put your answer in fully simplified point-slop

Archy [21]

$(\stackrel{x_1}{-6}~,~\stackrel{y_1}{-1})\qquad (\stackrel{x_2}{-4}~,~\stackrel{y_2}{2}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{2}-\stackrel{y1}{(-1)}}}{\underset{run} {\underset{x_2}{-4}-\underset{x_1}{(-6)}}} \implies \cfrac{2 +1}{-4 +6} \implies \cfrac{ 3 }{ 2 }$

$\begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{(-1)}=\stackrel{m}{\cfrac{3}{2}}(x-\stackrel{x_1}{(-6)}) \implies {\large \begin{array}{llll} y +1= \cfrac{3}{2} (x +6) \end{array}}$

4 0

1 year ago

On a coordinate plane, solid circles appear at the following points: (negative 3, 2), (negative 3, 3), (1, 4), (1, negative 4),

hammer [34]

Answer:

3 points

Step-by-step explanation:

A function is a relation that maps elements of a set for the inputs (the domain, usually denoted with the variable x) into elements of another set, the set of the outputs (the range, usually denoted with the variable y).

So the relation can be represented with points (x, y), which means that point x is being mapped into point y.

Such that the needed condition is that each element of the domain is mapped into only one element in the domain.

So for example, if for a relationship we have the points (a, b) and (a, c)

The point a is being mapped into two different outputs.

Now let's go to our problem, we have the points:

(-3, 2), (-3, 3), (1, 4), (1, -4), (2, -4), (2, -2)

If we have repeated inputs, we need to remove points until we have one of each.

We can see that the inputs -3, 1, and 2 are repeated one time.

Then we need to remove one of each, for example if we remove the second, fourth and sixth points, the set becomes:

(-3, 2), (1, 4), and (2, -4)

This is a function.

Then we need to remove 3 points.

7 0

3 years ago

Read 2 more answers