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2 years ago

R + 16 = 8 I need help please

2 answers:

4 0

ok so this is super simple! we take 16-8=8, so we cant use a positive number because we're adding in this situation so we add a negative number ( -8 + 16 = 8) so our solution R = -8

Lilit [14]2 years ago

3 0

Make the problem r=8-16. Then you will have r=-8, your answer

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The equation of a line XY is (y-3)=3/2(x-4) what is the slope of the line perpendicular to XY?

jok3333 [9.3K]

Answer:

It must be 3/2.........

3 0

2 years ago

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Alicia bikes 25% of a 100-mi trip on the first day. she bikes of the remaining distance on the second day. on the third day, she

DochEvi [55]

First day: 25% × 100 = 25 ( miles)
Second day : the remaining distance is 100 - 25 = 75 ( milesl
Third day: 40% × 75 = 30
25 + 75 + 30 = 130 (miles)

5 0

2 years ago

Use the intermediate value theorem to prove that there is a positive number c such that c2 = 2.

schepotkina [342]

So lets try to prove it,

So let's consider the function f(x) = x^2.

Since f(x) is a polynomial, then it is continuous on the interval (- infinity, + infinity).

Using the Intermediate Value Theorem,

it would be enough to show that at some point a f(x) is less than 2 and at some point b f(x) is greater than 2. For example, let a = 0 and b = 3.

Therefore, f(0) = 0, which is less than 2, and f(3) = 9, which is greater than 2. Applying IVT to f(x) = x^2 on the interval [0,3}.

Learn more about Intermediate Value Theorem on:

brainly.com/question/11377865

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6 0

1 year ago

The article on the next page from the San Francisco Chronicle, January 27, 1996, describes very serious problems of subsidence,

Mekhanik [1.2K]

Answer:

246.3%

the complete question is found in the attached document

Step-by-step explanation:

1st step:

using 1936 data,

w1= 356% = 3.56(356/100) , H= 79 feet

specific gravity = 2.65

Sₓ= 100%= 1

initial void ratio(e₀)= (w1 x specific gravity)/Sₓ

=3.56 x 2.65/1 = 9.434

2nd step

using 1996 data

ΔH= 22ft

ΔH/H = Δe/(1 + e₀)

22/79 = Δe/(1+9.434)

0.278=Δe/10.434

Δe= 0.278 x 10.434

Δe= 2.905

Δe= e₀ - eₓ

eₓ= e₀-Δe

eₓ= 9.434 - 2.905

eₓ= 6.529

3rd step

calculating water content in 1996

eₓ =6.529, specific gravity= 2.65, Sₓ= 100%

W2 X 2.65 = 1 x 6.529

w2 = 6.529/2.65 = 2.463 = 246.3%

6 0

3 years ago

A boat travels 12 km upstream and back in 1 hour 45 minutes. If the speed of the current is 3 km/h throughout, find the speed of

Elza [17]

Answer:

14.3 km/h to 3 significant figures

Step-by-step explanation:

Let time travelling upstream be t1 hours and downstream = t2 hours.

t1 + t2 = 1.75..........................(1)

Let v be the speed of the boat in still water.

12 / t1 = v - 3.........................(2)

12 / t2 = v + 3........................(3)

From (1) and (3) , substituting for t2:-

12 / (1.75 - t1) = v + 3 ..................(4)

Subtract:- (2) - (4):-

12 / t1 - 12 / (1.75 - t1) = -6

12(1 .75 - t1) - 12t1 = -6t1(1.75 - t1)

21 - 12t1 - 12t1 = -10.5t1 + 6t^2

6t1^2 + 13.5t1 - 21 = 0

Solving this gives t1 = 1.058 hours

Therefore the speed in still water (v) = 12/1.058 + 3 = 14.3 to 3 sig figs

3 0

2 years ago