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BigorU [14]
3 years ago
10

Josephina is four years younger than Cameron and the sum of their ages is 88. write an equation and solve it to figure out their

ages. How old are Josephina and Cameron?
Mathematics
2 answers:
slavikrds [6]3 years ago
3 0
Josephina is 42 and Cameron is 46. The equation is 42 plus 46 = 88.
WARRIOR [948]3 years ago
3 0
Josephina’s age = J
Cameron’s age = C

C-J = 4
C+J = 88

Solve for C in the first equation
C=4+J

Substitute
(4+J)+J=88
4+ 2J = 88
2J = 84
J = 42

Then plug that back in
C= 4+42
C=46

Josephina is 42 and Cameron is 46
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The height h, in feet, of a ball that is released 4 feet above the ground with an initial velocity of 80 feet per second is a fu
gulaghasi [49]

Answer with Step-by-step explanation:

The height of the ball from the ground as a function of time is given by

h(t)=-16t^2+80t+4

The height of the ball at any instant of time can be found by putting the value of time 't' in the above relation as

Part a)

Height of ball after 2 seconds it is released is

h(2)=-16\times 2^2+80\times 2+4=100feet

Part b)

Height of ball after 4 seconds it is released is

h(4)=-16\times 4^2+80\times 4+4=68feet

4 0
3 years ago
Read 2 more answers
1⁄2 × 1⁄5 =<br><br> help//.....
skad [1K]

Answer:

1/10

Step-by-step explanation:

1/2 * 1/5

Multiply the numerators

1*1 = 1

Multiply the denominators

2*5 =10

Numerator over denominator

1/10

7 0
2 years ago
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Determine the slope between (4, 2) and (-3, 5) ~hurry please~
SpyIntel [72]

Answer:

A.-3/7

Step-by-step explanation:

the slope formula is

m = (y2 - y1) / (x2-x1)

The point (4, 2) is (x1, y1) and the point (-3, 5) is (x2, y2). Now you have to substitute in the slope formula:

m = (5 - 2) / (-3 - 4)

m = 3 / -7

m = -3/7

7 0
3 years ago
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Find an equation of the tangent plane to the given parametric surface at the specified point.
Neko [114]

Answer:

Equation of tangent plane to given parametric equation is:

\frac{\sqrt{3}}{2}x-\frac{1}{2}y+z=\frac{\pi}{3}

Step-by-step explanation:

Given equation

      r(u, v)=u cos (v)\hat{i}+u sin (v)\hat{j}+v\hat{k}---(1)

Normal vector  tangent to plane is:

\hat{n} = \hat{r_{u}} \times \hat{r_{v}}\\r_{u}=\frac{\partial r}{\partial u}\\r_{v}=\frac{\partial r}{\partial v}

\frac{\partial r}{\partial u} =cos(v)\hat{i}+sin(v)\hat{j}\\\frac{\partial r}{\partial v}=-usin(v)\hat{i}+u cos(v)\hat{j}+\hat{k}

Normal vector  tangent to plane is given by:

r_{u} \times r_{v} =det\left[\begin{array}{ccc}\hat{i}&\hat{j}&\hat{k}\\cos(v)&sin(v)&0\\-usin(v)&ucos(v)&1\end{array}\right]

Expanding with first row

\hat{n} = \hat{i} \begin{vmatrix} sin(v)&0\\ucos(v) &1\end{vmatrix}- \hat{j} \begin{vmatrix} cos(v)&0\\-usin(v) &1\end{vmatrix}+\hat{k} \begin{vmatrix} cos(v)&sin(v)\\-usin(v) &ucos(v)\end{vmatrix}\\\hat{n}=sin(v)\hat{i}-cos(v)\hat{j}+u(cos^{2}v+sin^{2}v)\hat{k}\\\hat{n}=sin(v)\hat{i}-cos(v)\hat{j}+u\hat{k}\\

at u=5, v =π/3

                  =\frac{\sqrt{3} }{2}\hat{i}-\frac{1}{2}\hat{j}+\hat{k} ---(2)

at u=5, v =π/3 (1) becomes,

                 r(5, \frac{\pi}{3})=5 cos (\frac{\pi}{3})\hat{i}+5sin (\frac{\pi}{3})\hat{j}+\frac{\pi}{3}\hat{k}

                r(5, \frac{\pi}{3})=5(\frac{1}{2})\hat{i}+5 (\frac{\sqrt{3}}{2})\hat{j}+\frac{\pi}{3}\hat{k}

                r(5, \frac{\pi}{3})=\frac{5}{2}\hat{i}+(\frac{5\sqrt{3}}{2})\hat{j}+\frac{\pi}{3}\hat{k}

From above eq coordinates of r₀ can be found as:

            r_{o}=(\frac{5}{2},\frac{5\sqrt{3}}{2},\frac{\pi}{3})

From (2) coordinates of normal vector can be found as

            n=(\frac{\sqrt{3} }{2},-\frac{1}{2},1)  

Equation of tangent line can be found as:

  (\hat{r}-\hat{r_{o}}).\hat{n}=0\\((x-\frac{5}{2})\hat{i}+(y-\frac{5\sqrt{3}}{2})\hat{j}+(z-\frac{\pi}{3})\hat{k})(\frac{\sqrt{3} }{2}\hat{i}-\frac{1}{2}\hat{j}+\hat{k})=0\\\frac{\sqrt{3}}{2}x-\frac{5\sqrt{3}}{4}-\frac{1}{2}y+\frac{5\sqrt{3}}{4}+z-\frac{\pi}{3}=0\\\frac{\sqrt{3}}{2}x-\frac{1}{2}y+z=\frac{\pi}{3}

5 0
3 years ago
I dont know<br> How to solve this at all
AlladinOne [14]
So since this is a linear equation, which is y=mx+b, you use what it tells you in the word problem. so mx is the rate, since in this equation 200 per ton of sugar that's your mx, so its 200x. then for your b value, which in this case is the cost of truck rental, so its 5,400.

your equation is C=200s+5400


the variables are the same so you can use Y and X, but I used the variables used in the word problem.

Y=200x+5400
7 0
3 years ago
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