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jonny [76]
3 years ago
8

Write each fraction number as a decimal 33/40

Mathematics
1 answer:
Gnom [1K]3 years ago
3 0

Answer:

0.825

Step-by-step explanation:

In order to make the denominator 100, we must multiply both the numerator and denominator by 2.5

33*2.5 = 66 + 16.5 = 82.5

33/40 = 82.5/100

Now we solve, 82.5/100 = 0.825

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#19 4=ax - 14 Need asap
Vika [28.1K]

Answer

4 = ax - 14

18 = ax

x = 18/a or 4

Step-by-step explanation:

4 0
2 years ago
What is the distance between the sandbox and garden
elena55 [62]
For this case, the first thing we must do is apply the law of the sine.
 We then have the following expression:
 sine (65) / x = sine (104) / 36
 Clearing the value of x we have:
 x = (sine (65) / sine (104)) * (36)
 x = 33.6 feet
 Answer:
 
The distance between the sandbox and garden is:
 
x = 33.6 feet
6 0
3 years ago
A plant grows the same amount every week. Which graph matches the situation described? b e .​
dem82 [27]

Answer:

Option A

Step-by-step explanation:

Let the height of plant is 'b' units.

Graph representing the height of the plant will have y-intercept = b units

Since, the plant is growing at the same rate every week,

And growth of the plant is continuous.

Therefore, graph will be a straight line and continuous.

Since, the height of the plant is always increasing,

Slope of the line will be positive.

Option A will be the answer.

8 0
3 years ago
I am having trouble with number 2. Please show how the answer was gotten.
Andrews [41]

We know how we got 90, since a right angle is 90 degrees,

180 degrees are in a triangle.

180-90=90

Find the pair that equals 90 so you get 180 degrees!

If you have more questions don’t have hesitation to ask!

Or add all number to get 180 degrees!

4 0
3 years ago
Find the distance from the origin to the graph of 7x+9y+11=0
Cerrena [4.2K]
One way to do it is with calculus. The distance between any point (x,y)=\left(x,-\dfrac{7x+11}9\right) on the line to the origin is given by

d(x)=\sqrt{x^2+\left(-\dfrac{7x+11}9\right)^2}=\dfrac{\sqrt{130x^2+154x+121}}9

Now, both d(x) and d(x)^2 attain their respective extrema at the same critical points, so we can work with the latter and apply the derivative test to that.

d(x)^2=\dfrac{130x^2+154x+121}{81}\implies\dfrac{\mathrm dd(x)^2}{\mathrm dx}=\dfrac{260}{81}x+\dfrac{154}{81}

Solving for (d(x)^2)'=0, you find a critical point of x=-\dfrac{77}{130}.

Next, check the concavity of the squared distance to verify that a minimum occurs at this value. If the second derivative is positive, then the critical point is the site of a minimum.

You have

\dfrac{\mathrm d^2d(x)^2}{\mathrm dx^2}=\dfrac{260}{81}>0

so indeed, a minimum occurs at x=-\dfrac{77}{130}.

The minimum distance is then

d\left(-\dfrac{77}{130}\right)=\dfrac{11}{\sqrt{130}}
4 0
4 years ago
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