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romanna [79]
3 years ago
8

A worker earned $80 and 70% of it into the bank how many dollars did he put into the bank

Mathematics
2 answers:
Elenna [48]3 years ago
7 0
$56 goes into the bank, the equation would be 80 x .70 = 56
MA_775_DIABLO [31]3 years ago
3 0

Answer:

$56

Step-by-step explanation:

80 * .70 = 56

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Of the two bottles shown of sunblock shown, which is the better buy? Explain. Sunblock 1: 4 ounce bottle $8.49 Sunblock 2: 5 oun
Pepsi [2]

8.49/4 = 2.1225

Sunblock 1 costs $2.12 per ounce of sunblock.

10.29/5 = 2.058

Sunblock 2 costs $2.06 per ounce of sunblock.

Sunblock 2 is the better buy because it is cheaper ounce-per-ounce.

6 0
3 years ago
Compute the line integral with respect to arc length of the function f(x, y, z) = xy2 along the parametrized curve that is the l
SIZIF [17.4K]

Answer:

\displaystyle\frac{15\sqrt{3}}{4}-90\sqrt{146}

Step-by-step explanation:

The line integral with respect to arc length of the function f(x, y, z) = xy2 along the parametrized curve that is the line segment from (1, 1, 1) to (2, 2, 2) followed by the line segment from (2, 2, 2) to (−9, 6, 5) equals the sum of the line integral of f along each path separately.

Let  

C_1,C_2  

be the two paths.

Recall that if we parametrize a path C as (r_1(t),r_2(t),r_3(t)) with the parameter t varying on some interval [a,b], then the line integral with respect to arc length of a function f is

\displaystyle\int_{C}f(x,y,z)ds=\displaystyle\int_{a}^{b}f(r_1,r_2,r_3)\sqrt{(r'_1)^2+(r'_2)^2+(r'_3)^2}dt

Given any two points P, Q we can parametrize the line segment from P to Q as

r(t) = tQ + (1-t)P with 0≤ t≤ 1

The parametrization of the line segment from (1,1,1) to (2,2,2) is

r(t) = t(2,2,2) + (1-t)(1,1,1) = (1+t, 1+t, 1+t)

r'(t) = (1,1,1)

and  

\displaystyle\int_{C_1}f(x,y,z)ds=\displaystyle\int_{0}^{1}f(1+t,1+t,1+t)\sqrt{3}dt=\\\\=\sqrt{3}\displaystyle\int_{0}^{1}(1+t)(1+t)^2dt=\sqrt{3}\displaystyle\int_{0}^{1}(1+t)^3dt=\displaystyle\frac{15\sqrt{3}}{4}

The parametrization of the line segment from (2,2,2) to  

(-9,6,5) is

r(t) = t(-9,6,5) + (1-t)(2,2,2) = (2-11t, 2+4t, 2+3t)  

r'(t) = (-11,4,3)

and  

\displaystyle\int_{C_2}f(x,y,z)ds=\displaystyle\int_{0}^{1}f(2-11t,2+4t,2+3t)\sqrt{146}dt=\\\\=\sqrt{146}\displaystyle\int_{0}^{1}(2-11t)(2+4t)^2dt=-90\sqrt{146}

Hence

\displaystyle\int_{C}f(x,y,z)ds=\displaystyle\int_{C_1}f(x,y,z)ds+\displaystyle\int_{C_2}f(x,y,z)ds=\\\\=\boxed{\displaystyle\frac{15\sqrt{3}}{4}-90\sqrt{146}}

8 0
3 years ago
14. Of the 400 students in the 8th grade, four-
In-s [12.5K]

Answer:

8 or 9

Step-by-step explanation:

So, you first need to take 4/5 of 400 (320).  Then, you can count by 75s for a bit

75students; 2chaperones

150; 4

225; 6

300; 8

now, you have 20 students left over...since one chaperone would watch about 37 kids, you can either divide the remaining kids up among the other chaperones, or you can add one more chaperone to watch the 20 kids.  i don't know what the teacher wants, so either 8 or 9 chaperones

7 0
3 years ago
Read 2 more answers
Which function f (x) , graphed below, or g (x) , whose equation is g (x) = 3 cos 1/4 (x + x/3) + 2, has the largest maximum and
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Answer:

g(x), and the maximum is 5

Step-by-step explanation:

for given function f(x), the maximum can be seen from the shown graph i.e. 2

But for the function g(x), maximum needs to be calculated.

Given function :

g (x) = 3 cos 1/4 (x + x/3) + 2

let x=0 (as cosine is a periodic function and has maximum value of 1 at 0 angle)

g(x)= 3 cos1/4(0 + 0) +2

    = 3cos0 +2

     = 3(1) +2

     = 3 +2

     = 5 !

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Which expression makes the equation true for all values
Liula [17]
Can u show a picture of the problem
3 0
3 years ago
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