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2 years ago

Pleaseplease help me

Mathematics

1 answer:

Mars2501 [29]2 years ago

4 0

Answer:

1,4

Step-by-step explanation:

The anwser is 1,4

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Will Mark Brainlest helppp plss

bearhunter [10]

Answer:

? hey your question is not full at all

7 0

2 years ago

Ben has 50 drinks and paid 2$ for each. What is 67% of the final price ?

Debora [2.8K]

67% of final price is $67.

Step-by-step explanation:

Given,

Number of drinks Ben has = 50

Amount paid for each drink = $2

Amount paid for 50 drinks = 50*2

Amount paid for 50 drinks = final price = $100

67% of final price = $\frac{67}{100}*100$

67% of final price = $\frac{6700}{100}$

67% of final price = $67

67% of final price is $67.

Keywords: multiplication, percentage

Learn more about percentages at:

#LearnwithBrainly

7 0

2 years ago

The sum of partial products is equal to the final product?

Shkiper50 [21]

The sum of partial product is not always equal to a final product

8 0

3 years ago

Find the length x to the nearest whole number. A right triangle has a vertical leg of length x units, a base leg with a length o

wel

Complete Question

The diagram for this question is shown on the first uploaded image

Answer:

The value of x is $x = 274 \ unites$

Step-by-step explanation:

From the question we are told that

The length of the vertical length is $x$

The length of the base leg is L = 330 + d units

The length of the bisecting line segment is h

The base angle is $\theta = 28^o$

The angle between d and line segment is $\theta_1 = 56^o$

For the first angle

$Tan \theta_1 = \frac{x}{d}$

=> $d = \frac{x}{Tan \theta _1}$

For the whole big triangle

$Tan \theta = \frac{x}{330 + d}$

=> $d = \frac{x}{Tan \theta } -330$

So equating the both d

$\frac{x}{Tan \theta _1} = \frac{x}{Tan \theta } -330$

Substitute values

$\frac{x}{Tan (56)} = \frac{x}{Tan (28) } -330$

$0.6745 x = 1.880x -330$

$1.20549 x = 330$

$x = 274 \ unites$

5 0

2 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

2 years ago