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12345 [234]
4 years ago
15

45 is 15% of what number

Mathematics
2 answers:
blondinia [14]4 years ago
6 0
45 = 15% of number
45 = 15% × n
15% = \frac{15}{100} = 0.15

45 = 0.15n

\frac{45}{0.15} = \frac{0.15n}{0.15}
0.15 and 0.15 cancels out

300 = n

15% of 300 is 45

45 is 155 of 300
jekas [21]4 years ago
4 0
15% of a number is 45
15% × x = 45
x = 45 ÷ 15%
x = 45 ÷ 15/100
x = 45 × 100/15
x = 4500/15
x = 300

Check:
15% × 300
= 15/100 × 300
= 15 × 3
= 45

45 is 15% of 300.
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In right triangle DEF, ∠E is a right angle, m∠D=26∘, and DF=4.5.
jok3333 [9.3K]

Answer:

The measurement of EF is 1.98 units

Step-by-step explanation:

In Δ DEF

∵ m∠E = 90°

∴ FD is the hypotenuse of the triangle

∵ m∠D = 26°

∵ EF is opposite to angle D

- We can use the sine ratio to find it

∵ sin D = \frac{EF}{FD} ⇒ (opposite side / hypotenuse)

∵ FD = 4.5 units

∴ sin(26) = \frac{EF}{4.5}

- Multiply both sides by 4.5

∴ 4.5 sin(26) = EF

∵ sin 26° ≈ 0.44

∴ 4.5(0.44) = EF

∴ 1.98 = EF

The measurement of EF is 1.98 units

3 0
3 years ago
Plεαse solve thanks!!!!
Gwar [14]

Answer:

1/12

Step-by-step explanation:

1/6x

1/2

1/12

its lined up just in case

4 0
3 years ago
A cyclist rides at an average speed of 31 miles per hour. The cyclist’s speed, s, can differ from the average by as much as 8 mi
Shalnov [3]
In this compound inequality X should be 8.
4 0
3 years ago
Read 2 more answers
(10+ 2)12-06)<br> What is the answer?
Nutka1998 [239]
10+2=12-06/6=6
Hope this helped you ^^
8 0
3 years ago
Please help me on this problem
Anna71 [15]

Answer:

The pairs of integer having two real solution forax^{2} -6x+c = 0 are

  1. a = -4, c = 5
  2. a = 1, c = 6
  3. a = 2, c = 3
  4. a = 3, c = 3

Step-by-step explanation:

Given

ax^{2} -6x+c = 0

Now we will solve the equation by putting all the 6 pairs so we get the  following

-3x^{2} -6x-5 = 0 for a = -3 , c=-5

-4x^{2} -6x+5 = 0 for a = -4 , c=5

1x^{2} -6x+6 = 0 for a = 1 , c=6

2x^{2} -6x+3 = 0 for a = 2 , c=3

3x^{2} -6x+3 = 0 for a = 3 , c=3

5x^{2} -6x+4 = 0 for a = 5 , c=4

The above  all are Quadratic equations inn general form ax^{2} +bx+c=0

where we have a,b and c constant values

So for a real Solution we must have

Disciminant , b^{2} -4\timesa\timesc \geq 0

for a = -3 , c=-5 we have

Discriminant =-24 which is less than 0 ∴ not a real solution.

for a = -4 , c=5 we have

Discriminant = 116 which is greater than 0 ∴ a real solution.

for a = 1 , c=6 we have

Discriminant =12 which is greater than 0 ∴ a real solution.

for a = 2 , c=3 we have

Discriminant =12 which is greater than 0 ∴ a real solution.

for a = 3 , c=3 we have

Discriminant =0 which is equal to 0 ∴ a real solution.

for a = 5 , c=4 we have

Discriminant =-44 which is less than 0 ∴ not a real solution.

7 0
3 years ago
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