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3 years ago

Which of these is equivalent to -x < 8?: x < 8 x < -8 x > 8 x > -8

Mathematics

2 answers:

adell [148]3 years ago

6 0

Answer:

x>-8

Step-by-step explanation:

multiply both sides with -1

adoni [48]3 years ago

6 0

Answer:

Step-by-step explanation:

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Katena32 [7]

Answer:

300 m^2

Step-by-step explanation:

If you have any questions about the way I solved it, don't hesitate to ask =)

3 0

2 years ago

Find the points of intersection of the graphs of the equations.

AURORKA [14]

Answer:

To find the point of intersection, just solve for x and y in the equations

so since x - y = -5

x = y - 5

now that we have a value of x, we will substitute x with in in the other equation

5x + 3y = -9

5(y-5) + 3y = -9

5y - 25 + 3y = -9

8y = 16

y = 2

now use this value of y in the equation we made for x

x = y - 5

x = 2 - 5

x = -3

Hence, the point of contact is (-3 , 2)

4 0

3 years ago

The point (1/3,1/4) lies on the terminal said of an angle. Find the exact value of the six trig functions and explain which func

katrin2010 [14]

Answer:

sine and cosec are inverse of each other.

cosine and sec are inverse of each other.

tan and cot are inverse of each other.

Step-by-step explanation:

Given point on terminal side of an angle ( $\frac{1}{3},\frac{1}4$ ).

Kindly refer to the attached image for the diagram of the given point.

Let it be point A( $\frac{1}{3},\frac{1}4$ )

Let O be the origin i.e. (0,0)

Point B will be ( $\frac{1}{3},0$ )

Now, let us consider the right angled triangle $\triangle OBA$ :

Sides:

$Base, OB = \frac{1}{3}\\Perpendicular, AB = \frac{1}{4}$

Using Pythagorean theorem:

$\text{Hypotenuse}^{2} = \text{Base}^{2} + \text{Perpendicular}^{2}\\\Rightarrow OA^{2} = OB^{2} + AB^{2}\\\Rightarrow OA^{2} = \frac{1}{3}^{2} + \frac{1}{4}^{2}\\\Rightarrow OA = \sqrt{\frac{1}{3}^{2} + \frac{1}{4}^{2}}\\\Rightarrow OA = \sqrt{\frac{4^2+3^2}{3^{2}.4^2 }}\\\Rightarrow OA = \frac{5}{12}$

$sin \angle AOB = \dfrac{Perpendicular}{Hypotenuse}$

$\Rightarrow sin \angle AOB = \dfrac{\frac{1}{4}}{\frac{5}{12}}\\\Rightarrow sin \angle AOB = \dfrac{3}{5}$

$cos\angle AOB = \dfrac{Base}{Hypotenuse}$

$\Rightarrow cos \angle AOB = \dfrac{\frac{1}{3}}{\frac{5}{12}}\\\Rightarrow cos\angle AOB = \dfrac{4}{5}$

$tan\angle AOB = \dfrac{Perpendicular}{Base}$

$\Rightarrow tan\angle AOB = \dfrac{3}{4}$

$cosec \angle AOB = \dfrac{Hypotenuse}{Perpendicular}$

$\Rightarrow cosec\angle AOB = \dfrac{5}{3}$

$sec\angle AOB = \dfrac{Hypotenuse}{Base}$

$\Rightarrow sec\angle AOB = \dfrac{5}{4}$

$cot\angle AOB = \dfrac{Base}{Perpendicular}$

$\Rightarrow cot\angle AOB = \dfrac{4}{3}$

3 0

2 years ago

HELP ME PLEASE!

Delicious77 [7]

Sounds as tho' you have an isosceles triangle (a triangle with 2 equal sides). If this triangle is also a right triangle (with one 90-degree angle), then the side lengths MUST satisfy the Pythagorean Theorem.

Let's see whether they do.

8^2 + 8^2 = 11^2 ???
64 + 64 = 121? NO. This is not a right triangle.

If you really do have 2 sides that are both of length 8, and you really do have a right triangle, then:

8^2 + 8^2 = d^2, where d=hypotenuse. Then 64+64 = d^2, and

d = sqrt(128) = sqrt(8*16) = 4sqrt(8) = 4*2*sqrt(2) = 8sqrt(2) = 11.3.

11 is close to 11.3, but still, this triangle cannot really have 2 sides of length 8 and one side of length 11.

7 0

2 years ago

Mrs. Van and her family went bowling. The total cost of the shoes was $12 and the cost to bowl was $5 per game. Write an equatio

serious [3.7K]

y=5x+12

hope i could help!

5 0

2 years ago