Find the missing side lengths. Can someone help me please

Mathematics

1 answer:

Olenka [21]2 years ago

4 0

Those should be your answers I’m pretty sure

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estimate 0.00792398 to the nearest thousandth. Express your answer as a single-digit times a power of ten

Marianna [84]

Answer: $\approx8*10^{-3}$

Step-by-step explanation:

In order to round to the nearest thousandth, you must observe the digit to the right of the digit in the thousandth place and then:

If it is less than 5, you must round down.

If it is greater than or equal than 5, you must round up.

In this case, given the number:

$0.00792398$

You can notice that the digit to the right of the digit in the thousandth place is 9.

Since $9>5$ , you must round up. Then:

$0.00792398\approx0.008$

Finally, to express this as a single-digit times a power of ten, move the decimal point three places to the right. Then you get:

$\approx8*10^{-3}$

5 0

2 years ago

Use the distributive property for this expression: 6(2x+5y)

posledela

(6 times 2x) + (6 times 5y)

5 0

3 years ago

Find a quadratic polynomial with integer coefficients which has x = 2/9, ± SQRT23/9 as its real zeros.

melomori [17]

Answer:

The quadratic polynomial with integer coefficients is $y = 81\cdot x^{2}-36\cdot x -19$ .

Step-by-step explanation:

Statement is incorrectly written. Correct form is described below:

<em>Find a quadratic polynomial with integer coefficients which has the following real zeros: </em> $x = \frac{2}{9}\pm \frac{\sqrt{23}}{9}$ <em>. </em>

Let be $r_{1} = \frac{2}{9}+\frac{\sqrt{23}}{9}$ and $r_{2} = \frac{2}{9}-\frac{\sqrt{23}}{9}$ roots of the quadratic function. By Algebra we know that: