A mistake was made in the steps shown to simplify the expression

Stefan did 1 push-up on Sunday, 2 push-ups on Monday, 4 push-ups on Tuesday, 8 push-ups on Wednesday, and 16 push-ups on Thursda

statuscvo [17]

Answer:

32

Step-by-step explanation:

taste of change is by a factor of 2. it was 16 on Thursday. so 16x2 = 32

4 0

3 years ago

College Board

Hatshy [7]

<span>Since this is an SAT Math Level 2 problem derivatives should not be required to find the solution. To find "How many more hours of daylight does the day with max sunlight have than May 1," all you need to understand is that sin(x) has a maximum value of 1.

The day with max sunlight will occur when sin(2*pi*t/365) = 1, giving the max sunlight to be 35/3 + 7/3 = 14 hours

Evaluating your equation for sunlight when t = 41, May 1 will have about 13.18 hours of sunlight.

The difference is about 0.82 hours of sunlight.

Even though it is unnecessary for this problem, finding the actual max sunlight day can be done by solving for t when d = 14, of by the use of calculus. Common min/max problems on the SAT Math Level 2 involve sin and cos, which both have min values of -1 and max values of 1, and also polynomial functions with only even powered variables or variable expressions, which have a min/max when the variable or variable expression equals 0.

For example, f(x) = (x-2)^4 + 4 will have a min value of 4 when x = 2. Hope this helps</span>

3 0

3 years ago

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

Tasya [4]

Answer:

$\frac{dy}{dx}=-[(\frac{5x+24}{36x-6x^2})]$

Step-by-step explanation:

Given function:

y = $\ln(\frac{(6 - x)^{\frac{3}{2}}}{x^{\frac{2}{3}}})$

we know

$\ln(\frac{A}{B})$ = ln(A) - ln(B)

thus,

y = $\ln((6 - x)^{\frac{3}{2}}) - \ln(x^{\frac{2}{3}})$

or

also,

ln(Aⁿ) = n × ln(A)

thus,

y = $(\frac{3}{2})\times\ln(6 - x) - (\frac{2}{3})\times\ln(x)$

therefore,

$\frac{dy}{dx}=[(\frac{3}{2})\times\frac{1}{(6-x)}\times(0 - 1)] - [ (\frac{2}{3})\times\frac{1}{x}\times1]$

or

$\frac{dy}{dx}=-(\frac{3}{2(6-x)}) - (\frac{2}{3x})$

or

$\frac{dy}{dx}=-[(\frac{3(3x)+2\times2(6-x)}{2(6-x)\times(3x)})]$

or

$\frac{dy}{dx}=-[(\frac{5x+24}{36x-6x^2})]$

5 0

3 years ago

Please help me asap!

dusya [7]

7. Law of cosines

a² = b² + c² - 2bc cos A

where a is the side opposite to ∠A and b and c are adjacent sides to ∠A.

Here, a=BC= 3.8

b=CA =4.8

c = AB = 3.2

Substituting the values in the formula,

3.8² = 4.8² + 3.2² - 2(4.8)(3.2) cos A

30.72 cos A =23.04 + 10.24 -14.44=18.84

cos A = 18.84/30.72 = 0.6133

∴ cos A =0.6133

8. It is a right angled triangle.

Pythagoras theorem

hypotenuse² = Perpendicular² + base²

Here, hypotenuse = 6.7, base = 5.4

Perpendicular² = 6.7² -5.4²=44.89 - 29.16 =15.73

Perpendicular = √15.73 = 3.966 = 3.97

∴ The length of the missing side is 3.97.

9. Given: Side of new cube = 5 * Side of original cube

Let the side of the original cube measure 1 unit.

So, volume of original cube = side³ = 1

The side of new cube = 5 * 1 =5 units

So, volume of new cube = side³ = 5³ =125

The volume of the bigger cube is 125 times larger than the original cube.

10. Given : Height of the cone = 7 in

Radius of the base of the cone = 2 in

Volume of cone = (1/3) πr²h

=(1/3)π(2²)(7)

=29.32 cubic inches

∴ Volume of the cone is 29.32 cubic inches.

11. The given equation is:

x² -18x +10 =0

(x² -18x + 9²) +10 -9² =0

(x-9)²-71 =0

(x-9)² =71

x-9 =√71

x = 9 + √71

∴ x = 9+√71

7 0

3 years ago

This is 3 question help please

bekas [8.4K]

3 should be D
11 should be D
13 should be A

8 0

3 years ago