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4 years ago

30 POINT QUESTION!

Mathematics

2 answers:

kap26 [50]4 years ago

7 0

3-4 bathbombs and none

KengaRu [80]4 years ago

4 0

Answer: 3-4 thats all

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I’m really having trouble on this, can anyone help?

NikAS [45]

Answer:

the first box is proportional but the second is not

thus; y is (4/3)x for the first illustration

8 0

3 years ago

Plzzz help me I am timed plzzzzz

drek231 [11]

Your answer would be D or 6

3 0

3 years ago

PLEASE HELP PLEASE WILL GIVE BRAINILIEST 5 STARS AND THANK YOU TO WHOEVER CAN HELP!

Mnenie [13.5K]

Ok so the lines with two triangles are parralel to eachother. The lines with one are parallel to eachother.

The angle of 109° and angle of z° equal eachother. Z=109°

Since 109°, 33° and y° form a triangle, the sum of the angles will equal 180. Add 109 and 33 to get 142. Subtract 142 from 180 to get y°=38°.

Since z=109, this means that the triangle is congruent with the other. Since the congruent triangles are in a rhombus, then the angles are flipped. Thus, angle x =33°.

z=109°
y=38°
x=33°

3 0

3 years ago

What fraction of the months in the year have 30 days?

weeeeeb [17]

Answer:

4/12 which reduces to 1/3

Step-by-step explanation:

4 out of 12 months have 30 days.

8 0

3 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

3 years ago