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3 years ago

If school normally begins at 8:10 a.m. and we have a two hour delay, what time should we go to school?

Mathematics

2 answers:

Irina18 [472]3 years ago

6 0

Maybe 9.50 for prepare and reading some books

Burka [1]3 years ago

5 0

For this, you've just got to add 2 hours to 8:10, 8+2=10. Therefore, school would begin at 10:10am if there was a two hour delay :)

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A parking lot is 20 feet long. One tortoise starts at the edge of the parking lot and moves at a rate of 6 feet per minute. Anot

anygoal [31]

Answer:

Kindly check explanation

Step-by-step explanation:

Length of parking lot = 20 feets

Speed of tortoise which starts at the edge = 6 feets per minute

Speed of tortoise which starts 4 feets from the edge = 2 feets per minute

Equation to represent when they will be in the same spot.

Distance = speed * time

Distance of Tortoise at edge = 6ft/min * t = 6t - - (1)

Distance of the other tortoise = (4 + 2t) - - - (2)

Equating both (1) and (2)

6t = 4 + 2t

6t - 2t = 4

4t = 4

t = 4/4

t = 1

Hence, they'll be at the same spot after 1 minute.

6 0

3 years ago

The La Jolla tide pools are 67.5 miles from the SLMS. A map shows that 2 inches = 15 miles. What distance is SLMS from the tide

Gwar [14]

Answer:

Step-by-step explanation:

because you have to divide by 15's

5 0

3 years ago

Read 2 more answers

At what point does the curve have maximum curvature? Y = 4ex (x, y) = what happens to the curvature as x → ∞? Κ(x) approaches as

MAXImum [283]

<u>Answer-</u>

At $x= \frac{1}{2304e^4-16e^2}$ the curve has maximum curvature.

<u>Solution-</u>

The formula for curvature =

$K(x)=\frac{{y}''}{(1+({y}')^2)^{\frac{3}{2}}}$

Here,

$y=4e^{x}$

Then,

${y}' = 4e^{x} \ and \ {y}''=4e^{x}$

Putting the values,

$K(x)=\frac{{4e^{x}}}{(1+(4e^{x})^2)^{\frac{3}{2}}} = \frac{{4e^{x}}}{(1+16e^{2x})^{\frac{3}{2}}}$

Now, in order to get the max curvature value, we have to calculate the first derivative of this function and then to get where its value is max, we have to equate it to 0.

${k}'(x) = \frac{(1+16e^{2x})^{\frac{3}{2} } (4e^{x})-(4e^{x})(\frac{3}{2}(1+e^{2x})^{\frac{1}{2}})(32e^{2x})}{(1+16e^{2x} )^{2}}$

Now, equating this to 0

$(1+16e^{2x})^{\frac{3}{2} } (4e^{x})-(4e^{x})(\frac{3}{2}(1+e^{2x})^{\frac{1}{2}})(32e^{2x}) =0$

$\Rightarrow (1+16e^{2x})^{\frac{3}{2}}-(\frac{3}{2}(1+e^{2x})^{\frac{1}{2}})(32e^{2x})$

$\Rightarrow (1+16e^{2x})^{\frac{3}{2}}=(\frac{3}{2}(1+e^{2x})^{\frac{1}{2}})(32e^{2x})$

$\Rightarrow (1+16e^{2x})^{\frac{1}{2}}=48e^{2x}$

$\Rightarrow (1+16e^{2x})}=48^2e^{2x}=2304e^{2x}$

$\Rightarrow 2304e^{2x}-16e^{2x}-1=0$

Solving this eq,

we get $x= \frac{1}{2304e^4-16e^2}$

∴ At $x= \frac{1}{2304e^4-16e^2}$ the curvature is maximum.

6 0

2 years ago

(Brainliest) PLEASE HELPP MEEE

zloy xaker [14]

Answer:

1. According to the angle bisector theorem, DAC = BAC

2. Same kind of goes for BC

3. X must be equal in both equations. x= 3

9(3) -7 = 20

4(3) +8 = 20

Hope this helped!

8 0

2 years ago

Need help w precal/algebra 2

Daniel [21]

Answer:

g(-2) = -7

Step-by-step explanation:

To find g(-2), plug -2 into g(x).

g(x) = -4x² + 3x + 15

g(-2) = -4(-2)² + 3(-2) + 15

g(-2) = -4(4) + 3(-2) + 15

g(-2) = -16 + (-6) + 15

g(-2) = -1 + (-6)

g(-2) = -7

3 0

3 years ago