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3 years ago

!!!!!!!!!!!!help me

Mathematics

1 answer:

steposvetlana [31]3 years ago

5 0

Answer:

Step-by-step explanation:

i'll do a few of them

1) one inch plus a half inch plus an eighth inch plus a sixteenth inch

1 + 1/2 + 1/8 + 1/16 read the tick marks

1 + 8/16 + 2/16 + 1/16 find a common denominators

1 and 11/16 add the sixteenth numerator

4) 6 inch plus an eighth inch plus a sixteenth inch

6 + 1/8 + 1/16 read the tick marks

6 + 2/16 + 1/16 find a common denominators

6 and 3/16 add the sixteenth numerator

5) nine inch plus a quarter inch plus an eighth inch

9 + 1/4 + 1/8 read the tick marks

9 + 2/8 + 1/8 find a common denominators

9 and 3/8 add the sixteenth numerator

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In the graph above, the coordinates of the vertices of QPR are Q(3, 0), P(5, 6), and R(7, 0). If AQPR is reflected across

ozzi

Answer:

D. (7, 0)

Step-by-step explanation:

The rule for a reflection over the y-axis is (x, y) → (x, -y)

This means that the x-values stay the same while the y-values change.

Q(x, y) → (x, -y)

Q(3, 0) → (3, 0)

Q'(3, 0)

P(x, y) → (x, -y)

P(5, 6) → (5, -6)

P'(5, -6)

R(x, y) → (x, -y)

R(7, 0) → (7, 0)

R'(7, 0)

Therefore, the correct answer is D.

Hope this helps!

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2 years ago

What? I don’t under stand

s2008m [1.1K]

Answer:

? = h

Step-by-step explanation:

v = lwh

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v / wh = l

Hopefully this helped!

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5 0

3 years ago

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A man purchased 100 shares of a company’s stock at $3 per share. If each share rose $0.09 in value during the first month, decre

Neporo4naja [7]

$310 dollars because after it res and decreased it gained $.10 and 100 x 3 is 300 so thats before you calculate how much it rose so .10 x 100 is 10 so add that to 300 and you get 310

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2 years ago

Please help with this I am completely stuck on it

vaieri [72.5K]

Answer:

$f(x)=\sqrt[3]{x-4} , g(x)=6x^{2}\textrm{ or }f(x)=\sqrt[3]{x},g(x)=6x^{2} -4$

Step-by-step explanation:

Given:

The function, $H(x)=\sqrt[3]{6x^{2}-4}$

Solution 1:

Let $f(x)=\sqrt[3]{x}$

If $f(g(x))=H(x)=\sqrt[3]{6x^{2}-4}$ , then,

$\sqrt[3]{g(x)} =\sqrt[3]{6x^{2}-4}\\g(x)=6x^{2}-4$

Solution 2:

Let $f(x)=\sqrt[3]{x-4}$ . Then,

$f(g(x))=H(x)=\sqrt[3]{6x^{2}-4}\\\sqrt[3]{g(x)-4}=\sqrt[3]{6x^{2}-4} \\g(x)-4=6x^{2}-4\\g(x)=6x^{2}$

Similarly, there can be many solutions.

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3 years ago

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Setler79 [48]

Answer:

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Step-by-step explanation:

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3 years ago

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