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3 years ago

Over the weekend Ella ran three times as many miles as Vera. Together they ran 24 miles. How many miles did Vera run?

Mathematics

1 answer:

Butoxors [25]3 years ago

4 0

answer is 72

ella ran 3times as many and vera

24 x3=72 miles more

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The equation of a line is y = 2x + 3. What is the equation of the line that is parallel to the first line and passes through (2,

Masja [62]

Answer:

y = 2x - 5

Step-by-step explanation:

The slope for a parallel line is the same as the slope in the given equation. Therefore the slope of the new line will be 2 m= 2

use that slope and (2, -1) to find b

-1 = 2(2) + b

-1 = 4 + b

-5 = b

Plug m and b into y = mx + b

y = 2x - 5

8 0

3 years ago

Find the volume of the cone show below:

antiseptic1488 [7]

Formula for the volume not a cone=πr2H. π=3.14. R=12×2=24. H=16. 3.14×24=75.36. 75.36×16=1205.76. ANSWER: 1205.76

5 0

3 years ago

Differentiating a Logarithmic Function in Exercise, find the derivative of the function. See Examples 1, 2, 3, and 4.

adoni [48]

Answer: The required derivative is $\dfrac{8x^2+18x+9}{x(2x+3)^2}$

Step-by-step explanation:

Since we have given that

$y=\ln[x(2x+3)^2]$

Differentiating log function w.r.t. x, we get that

$\dfrac{dy}{dx}=\dfrac{1}{[x(2x+3)^2]}\times [x'(2x+3)^2+(2x+3)^2'x]\\\\\dfrac{dy}{dx}=\dfrac{1}{[x(2x+3)^2]}\times [(2x+3)^2+2x(2x+3)]\\\\\dfrac{dy}{dx}=\dfrac{4x^2+9+12x+4x^2+6x}{x(2x+3)^2}\\\\\dfrac{dy}{dx}=\dfrac{8x^2+18x+9}{x(2x+3)^2}$

Hence, the required derivative is $\dfrac{8x^2+18x+9}{x(2x+3)^2}$

3 0

3 years ago

Part 1

andreev551 [17]

Answer:

see explanation

Step-by-step explanation:

Part 1

The n th term of a geometric sequence is

$a_{n}$ = a $(r)^{n-1}$

where a is the first term and r the common ratio

Given $a_{6}$ = 128, then

a $(2)^{5}$ = 128, that is

32a = 128 ( divide both sides by 32 )

a = 4 ← first term

Part 2

$a_{21}$ = 4 $(2)^{20}$ = 4 × 1048576 = 4194304

6 0

3 years ago

Help me please ^^ much appreciated

Reil [10]

Answer:

Step-by-step explanation:

Using the sine ratio in the right triangle.

sinC = $\frac{opposite}{hypotenuse}$ = $\frac{AB}{BC}$ = $\frac{16}{28}$ , thus

∠ C = $sin^{-1}$ ( $\frac{16}{28}$ ) ≈ 34.85° ( to 2 dec. places )

8 0

3 years ago