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3 years ago

Explain how you can use base ten blocks to find 2.16 divide by 3

Mathematics

1 answer:

miskamm [114]3 years ago

3 0

You can use base ten blocks to find an answer by drwaing out models

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What is this expression in simplest form 1/2x^2-4x-2/x

Papessa [141]

Answer:

Step-by-step explanation:

Combine 12and x2.x22−4x−2x

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3 years ago

When an angle is reflected across a line, what happens to its measure

Vlad1618 [11]

It remains the same. if the shape is the same then so is it's angle measure.

7 0

3 years ago

Read 2 more answers

What is 1.38 as a fraction

slega [8]

<u>1.38</u> as a fraction is <em>138/100</em> .

As a fraction, it can be reduced to lower terms.

138/100 = 69 / 50

7 0

3 years ago

Read 2 more answers

7/8 -7/10 ,giving your answer as a fraction in its lowest terms

crimeas [40]

Answer:

7/40

Step-by-step explanation:

(70_56)/80

= 14/80

divided by 2

=7/40

7 0

3 years ago

On #10 i need help to solve this. please and thank you.

valina [46]

10a. $f(x) = x - 4$
   $h(x) = \sqrt{x - 5}$
   $(f - h)(6) = (6 - 4) - (\sqrt{6 - 5})$
   $(f - h)(6) = 2 - \sqrt{1}$
   $(f - h)(6) = 2 - 1$
   $(f - h)(6) = 1$

10b. $f(x) = x - 4$
   $g(x) = \frac{1}{x - 3}$
   $(f * g)(x) = (x - 4)(\frac{1}{x - 3})$
   $(f * g)(x) = \frac{x - 4}{x - 3}$
   Domain: (-∞, 3) ∨ (3, ∞) {x|x ≠ 3}
   Interval Notation: (3 < x < 3), (3 > x > 3)

10c. $g(x) = \frac{1}{x - 3}$
   $h(x) = \sqrt{x - 5}$
   $(g * h)(x) = (\frac{1}{x - 3})(\sqrt{x - 5})$
   $(g * h)(x) = \frac{\sqrt{x - 5}}{x - 3}$

10d. $f(x) = x - 4$
   $g(x) = \frac{1}{x - 3}$
   $h(x) = \sqrt[3]{x - 7}$
   $h(x) = (f * g)(x)$
   $\sqrt[3]{x - 7} = (x - 4)(\frac{1}{x - 3})$
   $\sqrt[3]{x - 7} = \frac{x - 4}{x - 3}$
   $(\sqrt[3]{x - 7})^{3} = (\frac{x - 4}{x - 3})^{3}$
   $x - 7 = \frac{(x - 4)^{3}}{(x - 3)^{3}}$
   $(x - 3)^{3}(x - 7) = (x - 4)^{3}$
   $(x^{3} - 9x^2 + 27x + 27)(x - 7) = (x^{3} - 12x^{2} + 48x - 64)$
   $x^{4} - 16x^{3} + 90x^{2} - 216x + 189 = x^{3} - 12x^{2} + 48x - 64$
   $x^{4} + 90x^{2} - 216x + 189 = 17x^{3} - 12x^{2} + 48x - 64$
   $x^{4} + 102x^{2} - 216x + 189 = 17x^{3} + 48x - 64$
   $x^{4} + 102x^{2} + 189 = 17x^{3} + 264x - 64$
   $x^{4} + 102x^{2} + 253 = 17x^{3} + 264x$
   $x^{4} - 17x^{3} + 102x^{2} - 264x + 253 = 0$
   $x = 4\ or\ 8$

4 0

3 years ago