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Ivahew [28]
2 years ago
13

PLZ PLZ PLZ WILL GIVE BRAINLIEST AND EXTRA POINTS!!! PLZ ANSWER!!! HEEEELLLP!!!!!! PLZ NO GO.OGLE!

Mathematics
1 answer:
Murrr4er [49]2 years ago
7 0

Answer:

48 / 4 = 12

use long division

       1 2

   4|48

     4 can fit into 4,  1 time

     4 can fit into 8,  2 times so you get 12

_______________________________

54 / 6 = 9

count how many times 6 goes into 54 witch is 9

_____________________________________________

plz mark me brainiest :)

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The following amounts were deposited in a savings account each month.
Ksju [112]

Answer: 20+44=64

              32+56=88

              88+64=$152

Step-by-step explanation:

3 0
3 years ago
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Look at pic 10 pts will mark brainilest
MissTica

Answer:

The answer is B or 8

Step-by-step explanation:

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2 years ago
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Which trigonometric functions are negative in the fourth (IV) quadrant?
anzhelika [568]
<h3>Answers are: sine, tangent, cosecant, cotangent</h3>

Explanation:

On the unit circle we have some point (x,y) such that x = cos(theta) and y = sin(theta). The sine corresponds to the y coordinate of the point on the circle. Quadrant IV is below the x axis which explains why sine is negative here, since y < 0 here.

Since sine is negative, so is cosecant as this is the reciprocal of sine

csc = 1/sin

In quadrant IV, cosine is positive as x > 0 here. So the ratio tan = sin/cos is going to be negative. We have a negative over a positive when we divide.

Because tangent is negative, so is cotangent.

The only positive functions in Q4 are cosine and secant, which is because sec = 1/cos.

7 0
2 years ago
Which is the equation of a hyperbola centered at the origin with x-intercept +\- 3 and asymptote y=2x
Radda [10]

Answer:

{\dfrac{x^{2}}{9} - \dfrac{y^{2}}{36} = 1}

Step-by-step explanation:

The hyperbola has x-intercepts, so it has a horizontal transverse axis.

The standard form of the equation of a hyperbola with a horizontal transverse axis is  \dfrac{(x - h)^{2}}{a^{2}} - \dfrac{(y - k)^{2}}{b^{2}} = 1

The center is at (h,k).

The distance between the vertices is 2a.

The equations of the asymptotes arey = k \pm \dfrac{b}{a}(x - h)

1. Calculate h and k. The hyperbola is symmetric about the origin, so  

h = 0 and k = 0

2. For 'a': 2a = x₂ - x₁ = 3 - (-3) = 3 + 3 = 6

a = 6/2 = 3  

3. For 'b': The equation for the asymptote with the positive slope is  

y = k + \dfrac{b}{a}(x - h) = \dfrac{b}{a}x

Thus,  asymptote has the slope of

\begin{array}{rcl}m& =& \dfrac{b}{a}\\\\2& =& \dfrac{b}{3}\\\\b& =& \mathbf{6}\end{array}

4.  The equation of the hyperbola is

\large \boxed{\mathbf{\dfrac{x^{2}}{9} - \dfrac{y^{2}}{36} = 1}}

The attachment below represents your hyperbola with x-intercepts at ±3 and asymptotes with slope ±2.

7 0
2 years ago
The rectangular prism shown has a length of 5 cm, a height of 3 cm, and the width of 4 cm. What is the surface area of this pris
PilotLPTM [1.2K]

Answer:20

Step-by-step explanation:

because you are finding out the base so width times length

8 0
2 years ago
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