All categories

2 years ago

1plss help, its time

Mathematics

2 answers:

DanielleElmas [232]2 years ago

8 0

Answer:

60 degrees

Step-by-step explanation:

Similar shapes have the same angles but different side lengths.

bagirrra123 [75]2 years ago

7 0

Answer:

The answer is 60°

Step-by-step explanation:

Because pentagon JKLMN is basically the same as pentagon VWXYZ and if you see in the smaller pentagon the angle for L is 90° and same goes for the bigger pentagon

You might be interested in

¿Cómo las razones de seno y cos<br>eno son semejantes?

Sauron [17]

Las razones de los lados de un triángulo rectángulo se llaman razones trigonométricas.

Espero te ayude :)

6 0

2 years ago

Solve the following equation. 3x/7-2=15 (Must be a mixed number)

Elina [12.6K]

Answer:

Mixed Number

$39 \frac{2}{3}$

4 0

3 years ago

kenny6666 [7]

<span>△PQR is similar to △STU

</span>m∠R = m∠U = 96°

m∠Q = m∠T = 6

m∠P = 180 - ( 96 + 6)

m∠P = 180 - 102

m∠P = 78

answer

m∠P = 78

7 0

3 years ago

Read 2 more answers

consider the following equations: -x-y=1 y=x+3 if the two equations are graphed, at what point do the lines representing the two

Oksana_A [137]

Answer:

(-2, 1)

Step-by-step explanation:

consider the following equations: -x-y=1 y=x+3

For the lines to intersect, the values must be the same

-x-y=1

x + y = -1

y = -x - 1

Equating the y values

-x - 1 = x+ 3

-x-x = 3 + 1

-2x = 4

x = 4/-2

x = -2

Substitute x = -2 into y = x+3

y = -2 + 3

y = 1

Hence the required coordinate is (-2, 1)

7 0

2 years ago

<img src="https://tex.z-dn.net/?f=%28x%5E%7B2%7D%20%2Bx-3%29%3A%20%28x%5E%7B2%7D%20-4%29%5Cgeq%201" id="TexFormula1" title="(x^{

Jlenok [28]

Answer:

$x>2$

Step-by-step explanation:

When given the following inequality;

$(x^2+x-3):(x^2-4)\geq1$

Rewrite in a fractional form so that it is easier to work with. Remember, a ratio is another way of expressing a fraction where the first term is the numerator (value over the fraction) and the second is the denominator(value under the fraction);

$\frac{x^2+x-3}{x^2-4}\geq1$

Now bring all of the terms to one side so that the other side is just a zero, use the idea of inverse operations to achieve this:

$\frac{x^2+x-3}{x^2-4}-1\geq0$

Convert the (1) to have the like denominator as the other term on the left side. Keep in mind, any term over itself is equal to (1);

$\frac{x^2+x-3}{x^2-4}-\frac{x^2-4}{x^2-4}\geq0$

Perform the operation on the other side distribute the negative sign and combine like terms;

$\frac{(x^2+x-3)-(x^2-4)}{x^2-4}\geq0\\\\\frac{x^2+x-3-x^2+4}{x^2-4}\geq0\\\\\frac{x+1}{x^2-4}\geq0$

Factor the equation so that one can find the intervales where the inequality is true;

$\frac{x+1}{(x-2)(x+2)}\geq0$

Solve to find the intervales when the equation is true. These intervales are the spaces between the zeros. The zeros of the inequality can be found using the zero product property (which states that any number times zero equals zero), these zeros are as follows;

$-1, 2, -2$

Therefore the intervales are the following, remember, the denominator cannot be zero, therefore some zeros are not included in the domain

$x\leq-2\\-2$

Substitute a value in these intervales to find out if the inequality is positive or negative, if it is positive then the interval is a solution, if it is negative then it is not a solution. This is because the inequality is greater than or equal to zero;

$x\leq-2$ -> negative

$-2$ -> neagtive

$-1\leq x$ -> neagtive

$x>2$ -> positive

Therefore, the solution to the inequality is the following;

$x>2$

6 0

2 years ago