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2 years ago

What are the coordinates of the vertex of the graph:

Mathematics

2 answers:

Aleks04 [339]2 years ago

8 0

Answer is C. (-7, -16)

fgiga [73]2 years ago

5 0

Answer:

(-7,-16)

Step-by-step explanation:

calculate the first derivative and equale tot zero.

dy/dx = 2x + 14 = 0

<=> x = -7

calculate now the value of y for x = -7

y = (-7)^2 + 14*(-7) + 33

y = 49 - 98 + 33

y = -16

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Help me with these math questions....

Nady [450]

Answer: cotθ

<u>Step-by-step explanation:</u>

tanθ * cos²θ * csc²θ

= $\dfrac{sin\theta}{cos\theta} * \dfrac{cos\theta*cos\theta}{} *\dfrac{1}{sin\theta*sin\theta}$

= $\dfrac{cos\theta}{sin\theta}$

= cotθ

Answer: B

<u>Step-by-step explanation:</u>

The parent graph is y = x²

The new graph y = -x² + 3 should have the following:

reflection over the x-axis
vertical shift up 3 units

Answers:

a. Quadrant II
b. negative
c. $\dfrac{\pi}{6}$
d. C
e. $-\dfrac{\sqrt{3}}{3}$

<u>Explanation:</u>

$\dfrac{17\pi}{6} - \dfrac{12\pi}{6} = \dfrac{5\pi}{6}$

a) Quadrant 2 is: $\dfrac{\pi}{2} < \theta < \pi$

b) In Quadrant 2, cos is negative and sin is positive, so tan is negative

c) $\pi-\dfrac{5\pi}{6}$ = $\dfrac{\pi}{6}$

d) the reference line is above the x-axis so it is negative --> $-tan\dfrac{\pi}{6}$

e) $tan(\dfrac{5\pi}{6})=\dfrac{1}{-\sqrt{3}}=-\dfrac{\sqrt{3}}{3}$

4 0

3 years ago

Find the numbers of faces, edges, and vertices of the solid.

deff fn [24]

Answer:

7 faces
15 edges
10 vertices

Step-by-step explanation:

This is a counting problem. As with many counting problems, it is helpful to adopt a strategy that helps ensure you count everything only once.

<h3>Faces</h3>

There are two pentagonal faces and 5 rectangular faces for a total of ...

7 faces

<h3>Edges</h3>

There are 5 edges around each of the pentagonal faces, and 5 edges connecting the top face to the bottom faces, for a total of ...

15 edges

<h3>Vertices</h3>

There are 5 vertices on the top face, and 5 on the bottom face, for a total of ...

10 vertices

7 0

2 years ago

Quadrilateral PQRS is a square whos side length is 10. Let X and Y be points outside the square so that XQ = YS = 6 and XP = YR

erik [133]

Answer:

392

Step-by-step explanation:

Triangles XQP and YRS are right triangles because triples 6, 8, 10 are Pythagorean triples.

Extend lines XQ, YR, YS and XP and mark their intersection as A and B.

Quadrilateral XAYB is a square because all right triangles PXQ, QAR, RYS and SBP are congruent (by ASA postulate) and therefore

all angles of the quadrilateral XAYB are right angles

all sides of XAYB are congruent and equal to 6 + 8 = 14 units.

Segment XY is the diagonal of the square XAYB, by Pythagorean theorem,

$XY^2=XA^2+AY^2\\ \\XY^2=14^2+14^2\\ \\XY^2=196+196\\ \\XY^2=392$

3 0

3 years ago

Please someone help me

Yuki888 [10]

Answer: H

Step-by-step explanation:

We can see this rectangular prism will have side lengths of 5, 3, 7.

So we can use the fact Surface Area = 2(lh+wh+lw) to get:

2(15 + 35 + 21) = 2(71) = 142

5 0

2 years ago

Sketch the image of quadrilateral PQRS under the following dilations:

kramer

Answer:

dam this been up for a long time

Step-by-step explanation:

its just too hard sorry man

7 0

3 years ago