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torisob [31]
3 years ago
9

Danny had 20 minutes to do a three-problem quiz. He spent 9 1/4 minutes on question A and 5 4/5 minutes on question B. How much

time did he have left for question C?
Mathematics
1 answer:
Travka [436]3 years ago
6 0

Answer:

4 19/20

Step-by-step explanation:

9 1/4 + 5 4/5 = 15 1/20

20- 15 1/20 = 4  19/20

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Which values are solutions to the inequality below? √x ≤ 5
Romashka [77]
\sqrt{x}  \leq 5

Find the real  region  for \sqrt{x}

\sqrt{x} is real for x \geq 0

= x  \geq 0

\sqrt{x}  \leq 5

Square both sides:

( \sqrt{x} )^2  \leq 5^2

5^2 = 5*5 =25

Refine:

x  \leq 25

Answer D

hope this helps!


5 0
3 years ago
Which number is the largest? 7.2 ⋅ 10−6, 3.09 ⋅ 103, 2.04 ⋅ 104, 5 ⋅ 103
Tems11 [23]
The equation whit the largest outcome is 5*103=515
7 0
3 years ago
Read 2 more answers
5(2x)to the third power
kumpel [21]
(5(2x))^3
Use the property of exponents:
(5^3)*(2x)^3
Simplify the exponents:
125*8*x^3
Simplify the numbers and multiply the integers:
1000*x^3

Hope this helps :)
5 0
3 years ago
Which statement accurately describes how to reflect point A(3,-1) over the y-axis?
Lyrx [107]

Answer:

The answer in the procedure

Step-by-step explanation:

we know that

The rule of the reflection of a point over the y-axis is equal to

A(x,y) ----->A'(-x,y)

That means -----> The x-coordinate of the image is equal to the x-coordinate of the pre-image multiplied by -1 and the y-coordinate of both points (pre-image and image) is the same

so

A(3,-1) ------> A'(-3,-1)

The distance from A to the y-axis is equal to the distance from A' to the y-axis (is equidistant)

therefore

To reflect a point over the y-axis

Construct a line from A perpendicular to the y-axis, determine the distance from A to the y-axis along this perpendicular line, find a new point on the other side of the y-axis that is equidistant from the y-axis

6 0
3 years ago
The diagonal of a rectangle is of length a. It splits each corner forming two angles with a ratio of 1:2. The area of the rectan
HACTEHA [7]

Answer:

Step-by-step explanation:

Given

Length of diagonal is a

Diagonal divides the angle in 1:2

such that \theta +2\theta =90 (because angle between two sides is 90)

3\theta =90

\theta =30^{\circ}

width of rectangle is b=a\sin \theta =\frac{a}{2}

Length of rectangle is L=a\cos 30=\frac{\sqrt{3}}{2}a

Area of rectangle A=L\cdot b

A=\frac{\sqrt{3}}{2}a\times \frac{a}{2}

A=\frac{\sqrt{3}}{4}a^2

5 0
3 years ago
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