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3 years ago

Can somebody plz help me

Mathematics

1 answer:

Dominik [7]3 years ago

8 0

This type of proof is called a flowchart proof.

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Find the value of (2^−2 − 4^0) × 4^−2

Dafna1 [17]

Answer:

-0,046875

Step-by-step explanation:

$( {2}^{ - 2} - {4}^{0} ) \times {4}^{ - 2} \\ = (0.25 - 1) \times 0.0625 \\ = - 0.75 \times 0.0625 \\ = - 0.046875$

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2 years ago

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Pls help ASAP

soldier1979 [14.2K]

Answer:

Sample response:

To determine the relationship between quantities, you must determine what to do to the x-values to make them into y-values. The correct operation must turn every x-value into the corresponding y- value in the table. Once you know the relationship, you can use the same operation on all of the x-values that have unknown y-values.

OR:

In a table to get from x to y you have to multiply by a number. To find this you must divide x by y or y by x. So to get the missing quantities you must multiply the last known number by the number you found by dividing.

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3 years ago

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Solve for y. 9x+5y=7 <br>

Yuliya22 [10]

Answer: y=7/5-9x/5

Step-by-step explanation:

im sorry if this dont help

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2 years ago

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What is the relationship between 5,000 and 500,000

solong [7]

5000 is 1/100 of the value of 500,000

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3 years ago

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Help pleaseeeeeeeeeeeeeeeeeeeeeee

bixtya [17]

Answer: $\bold{(1)\ \dfrac{19,683}{64}\qquad (2)\ 16}$

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

(1) (12, 18, 27, ...)

The common ratio is:

$r=\dfrac{a_{n+1}}{a_n}\quad r =\dfrac{18}{12}=\boxed{\dfrac{3}{2}}\quad \rightarrow \quad r=\dfrac{27}{18}=\boxed{\dfrac{3}{2}}$

The equation is:

$a_n=a_o(r)^{n-1}\\\\Given:a_o=12,\ r=\dfrac{3}{2}\\\\\\Equation:\\a_n =12\bigg(\dfrac{3}{2}\bigg)^{n-1}\\\\\\\\9th\ term:\\a_9=12\bigg(\dfrac{3}{2}\bigg)^{9-1}\\\\\\a_9=12\bigg(\dfrac{3}{2}\bigg)^{8}\\\\\\.\quad =\large\boxed{\dfrac{19643}{64}}$

$(2)\qquad \bigg(\dfrac{1}{16},\dfrac{1}{8},\dfrac{1}{4},\dfrac{1}{2}\bigg)\\\\\\\text{The common ratio is}:\\\\r=\dfrac{a_{n+1}}{a_n}\quad r=\dfrac{\frac{1}{8}}{\frac{1}{16}}=\boxed{2}\quad \rightarrow \quad r=\dfrac{\frac{1}{4}}{\frac{1}{8}}=\boxed{2}$

The equation is:

$a_n=a_o(r)^{n-1}\\\\Given:a_o=\dfrac{1}{16},\ r=2\\\\\\Equation:\\a_n =\dfrac{1}{16}(2)^{n-1}\\\\\\\\9th\ term:\\a_9=\dfrac{1}{16}(2)^{9-1}\\\\\\a_9=\dfrac{1}{16}(2)^{8}\\\\\\.\quad =\large\boxed{16}$

3 0

3 years ago