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Paul [167]
3 years ago
9

Need help with this question

Mathematics
1 answer:
podryga [215]3 years ago
5 0

Answer:

The bottom two

Step-by-step explanation:

Units of time should always be on the x-axis, otherwise a positive graph would mean a drop in the opposing y-axis unit.

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What is the value of x? Enter your answer in the box. x = °
Elodia [21]
Answer: 90°
Explanation: the sum of all the angles in a triangle = 180. So 70 + 30 = 90 and 180-90=90
3 0
3 years ago
In ΔPQR, the measure of ∠R=90°, the measure of ∠Q=46°, and RP = 97 feet. Find the length of PQ to the nearest tenth of a foot.
Whitepunk [10]

Answer:

PQ = 134.9 feet

Step-by-step explanation:

sin 46 = 97/PQ

0.7193 = 97/PQ

PQ = 134.85 feet

5 0
3 years ago
Let f(x) be a continuous function such that f(1) and f'(x)-Vx3 + 6. What is the value of,f(5)? (A) 11.446 (C) 24.672 (B) 13.446
tia_tia [17]

Answer:

f(5) = 26.672 which is option D

Step-by-step explanation:

From question, f(1) = 2 and f'(x)=√(x^3 + 6)

f(5) = f(1) + (5,1)∫ f'(x) dx

Integrating using the boundary 5 and 1;

f(5) = 2 + (5,1)∫√(x^3 + 6) dx

f(5) = 2 + 24.672

So f(5) = 26.672

7 0
3 years ago
Segment KJ shown below is the hypotenuse of isosceles right triangle JLK.
irga5000 [103]
On a coordinate plane it is negetive 5
7 0
3 years ago
Read 2 more answers
Let S be the surface defined by x 2 + 2y 3 + 3z 4 = 6. Let T be the surface defined parametrically by r(u, v) = (1+ln u, 2e v+u−
aleksandrvk [35]

The tangent to C through (1, 1, 1) must be perpendicular to the normal vectors to the surfaces S and T at that point.

Let f(x,y,z)=x^2+2y^3+3z^4. Then S is the level curve f(x,y,z)=6. Recall that the gradient vector is perpendicular to level curves; we have

\nabla f(x,y,z)=(2x,6y,12z^2)

so that the gradient of f at (1, 1, 1) is

\nabla f(1,1,1)=(2,6,12)

For the surface T, we have

\begin{cases}1+\ln u=1\\2e^v+u-2=1\\uv+1=1\end{cases}\implies u=1,v=0

so that \vec r(1,0)=(1,1,1). We can obtain a vector normal to T by taking the cross product of the partial derivatives of \vec r(u,v), and evaluating that product for u=1,v=0:

\dfrac{\partial\vec r}{\partial u}\times\dfrac{\partial\vec r}{\partial v}=\left(u-2ve^v,-1,\dfrac{2e^v}u\right)

\left(\dfrac{\partial\vec r}{\partial u}\times\dfrac{\partial\vec r}{\partial v}\right)(1,0)=(1,-1,2)

Now take the cross product of the two normal vectors to S and T:

(2,6,12)\times(1,-1,2)=(24,8,-8)

The direction of vector (24, 8, -8) is the direction of the tangent line to C at (1, 1, 1). We can capture all points on the line containing this vector by scaling it by t\in\Bbb R. Then adding (1, 1, 1) shifts this line to the point of tangency on C. So the tangent line has equation

\vec\ell(t)=(1,1,1)+t(24,8,-8)=(1+24t,1+8t,1-8t)

7 0
3 years ago
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