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3 years ago

CAN SOMEONE HELP WITH THE QUESTION IN THE IMAGE.

Mathematics

1 answer:

levacccp [35]3 years ago

6 0

Answer:

8.4

Step-by-step explanation:

Since the four horizontal lines are parallel, they each form congruent corresponding angles with the transversal line SP. Since PQ = QR = RS, that means the vertical distances between the parallel lines is equal.

Since the parallel lines are equally spaced, and form congruent angles with the transversal line TU, then TX = XZ = ZU = 2.8.

So the length of TU is 2.8 + 2.8 + 2.8 = 8.4.

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Number 5 pleaseeeeee

Korvikt [17]

The answer is C. y= 3/2 x. You can check by multiplying 3/2 by 3, if you get 7, it is correct

6 0

3 years ago

Write an equation in slope intercept form that passes through the point (3,9) and has a slope of -5

Anna35 [415]

Answer:

y= -5x +24

Step-by-step explanation:

<u>Slope-intercept form</u>

y= mx +c, where m is the slope and c is the y-intercept.

Given that the slope is -5, m= -5.

Substitute m= -5 into the equation:

y= -5x +c

To find the value of c, substitute a pair of coordinates that the line passes through into the equation:

When x= 3, y= 9,

9= -5(3) +c

9= -15 +c

c= 9 +15

c= 24

Thus, the equation of the line is y= -5x +24.

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2 years ago

Can you the nearest hundredth to 1.2983?

ozzi

I think so it’s 1.30

3 0

2 years ago

Read 2 more answers

Suppose that \nabla f(x,y,z) = 2xyze^{x^2}\mathbf{i} + ze^{x^2}\mathbf{j} + ye^{x^2}\mathbf{k}. if f(0,0,0) = 2, find f(1,1,1).

lesya [120]

The simplest path from (0, 0, 0) to (1, 1, 1) is a straight line, denoted $C$ , which we can parameterize by the vector-valued function,

$\mathbf r(t)=(1-t)(\mathbf i+\mathbf j+\mathbf k)$

for $0\le t\le1$ , which has differential

$\mathrm d\mathbf r=-(\mathbf i+\mathbf j+\mathbf k)\,\mathrm dt$

Then with $x(t)=y(t)=z(t)=1-t$ , we have

$\displaystyle\int_{\mathcal C}\nabla f(x,y,z)\cdot\mathrm d\mathbf r=\int_{t=0}^{t=1}\nabla f(x(t),y(t),z(t))\cdot\mathrm d\mathbf r$

$=\displaystyle\int_{t=0}^{t=1}\left(2(1-t)^3e^{(1-t)^2}\,\mathbf i+(1-t)e^{(1-t)^2}\,\mathbf j+(1-t)e^{(1-t)^2}\,\mathbf k\right)\cdot-(\mathbf i+\mathbf j+\mathbf k)\,\mathrm dt$