Ms.Gallegos burns 236 calories

Problem page scott has scored 28, 22, and 29 points in his three basketball games so far. how many points does he need to score

diamong [38]

28+22+29+ 21 (your answer) = 100 divided by 4 (the amount of numbers used) = 25 (average)

3 0

3 years ago

Last Monday local mail carriers delivered 1,344 pieces of mail. Of all the deliveries they made, 1 / 6 were magazines, 1 / 12 we

Brut [27]

Answer: 1008 letters

Step-by-step explanation:

From the question, last Monday, a local mail carriers delivered 1,344 pieces of mail and of all the deliveries they made, 1 / 6 were magazines, 1 / 12 were packages, and the rest were letters. The fraction of letters will be:

= 1 - (1/6 + 1/12)

= 1 - (3/12)

= 9/12 = 3/4

We will now multiply 3/4 by 1344 to get the number of letters delivered.

= 3/4 × 1344

= 1008 letters

6 0

3 years ago

Which of the following is a trinomial?

Sedaia [141]

Answer:

$A) c^2 + c + 6$

Step-by-step explanation:

We have been given four choices out of which we need to select the choice which is trinomial.

Trinomial means polynomial having three terms.

A) c2 + c + 6

It has 3 terms.

B) c2 − 16

It has 2 terms.

C) −8c

It has 1 term.

D) c3 + 4c2 − 12c + 7

It has 4 terms.

We see that only choice A) has three terms.

Hence correct choice is $A) c^2 + c + 6$

5 0

3 years ago

a circle with a radius of 1 cm sits inside a 3cm x 3cm rectangle. What is the area of the shaded region (the rectangle is shaded

masha68 [24]

Answer:

The area of the shaded region is $5.86\ cm^{2}$

Step-by-step explanation:

we know that

The area of the shaded region is equal to the area of the rectangle minus the area of the circle

The rectangle is a square

so

$A=b^{2} -\pi r^{2}$

we have

$b=3\ cm$

$r=1\ cm$

assume

$\pi =3.14$

substitute the values

$A=3^{2} -(3.14)(1)^{2}$

$A=5.86\ cm^{2}$

5 0

3 years ago

Read 2 more answers

Find the length of the curve. R(t) = cos(8t) i + sin(8t) j + 8 ln cos t k, 0 ≤ t ≤ π/4

arsen [322]

we are given

$R(t)=cos(8t)i+sin(8t)j+8ln(cos(t))k$

now, we can find x , y and z components

$x=cos(8t),y=sin(8t),z=8ln(cos(t))$

Arc length calculation:

we can use formula

$L=\int\limits^a_b {\sqrt{(x')^2+(y')^2+(z')^2} } \, dt$

$x'=-8sin(8t),y=8cos(8t),z=-8tan(t)$

now, we can plug these values

$L=\int _0^{\frac{\pi }{4}}\sqrt{(-8sin(8t))^2+(8cos(8t))^2+(-8tan(t))^2} dt$

now, we can simplify it

$L=\int _0^{\frac{\pi }{4}}\sqrt{64+64tan^2(t)} dt$

$L=\int _0^{\frac{\pi }{4}}8\sqrt{1+tan^2(t)} dt$

$L=\int _0^{\frac{\pi }{4}}8\sqrt{sec^2(t)} dt$

$L=\int _0^{\frac{\pi }{4}}8sec(t) dt$

now, we can solve integral

$\int \:8\sec \left(t\right)dt$

$=8\ln \left|\tan \left(t\right)+\sec \left(t\right)\right|$

now, we can plug bounds

and we get

$=8\ln \left(\sqrt{2}+1\right)-0$

so,

$L=8\ln \left(1+\sqrt{2}\right)$ ..............Answer

5 0

3 years ago