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2 years ago

The lesser of two consecutive even integers is 10 more than one-half the greater.Find the integers

Mathematics

1 answer:

Flauer [41]2 years ago

8 0

The two consecutive integers are 21 and 22.

<h3>How to find the two consecutive integers?</h3>

Two consecutive integer numbers can be written as x and x + 1.

The lesser of these two numbers is x, and we know that it is 10 more than one-half of the greater integer, then we can write:

x = 10 + (x + 1)/2

Now we can solve that equation for x:

x = 10 + (x + 1)/2

2x = 20 + (x + 1)

2x = 21 + x

x = 21

Then the lesser number is x = 21 and the larger number is x + 1 = 22

If you want to learn more about integer numbers:

brainly.com/question/2493629

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ohaa [14]

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3 years ago

The product of 6 and the total of z and 9

tatiyna

Answer:

6(z+9)

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Step-by-step explanation:

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2 years ago

8 / 3 3/8 as a mixed number in simplest form

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2 years ago

The population of Henderson City was 3,381,000 in 1994, and is growing at an annual rate 1.8%

liq [111]

<h2>In the year 2000, population will be 3,762,979 approximately. Population will double by the year 2033.</h2>

Step-by-step explanation:

Given that the population grows every year at the same rate( 1.8% ), we can model the population similar to a compound Interest problem.

From 1994, every subsequent year the new population is obtained by multiplying the previous years' population by $\frac{100+1.8}{100}$ = $\frac{101.8}{100}$ .

So, the population in the year t can be given by $P(t)=3,381,000\textrm{x}(\frac{101.8}{100})^{(t-1994)}$

Population in the year 2000 = $3,381,000\textrm{x}(\frac{101.8}{100})^{6}$ = $3,762,979.38$

Population in year 2000 = 3,762,979

Let us assume population doubles by year $y$ .

$2\textrm{x}(3,381,000)=(3,381,000)\textrm{x}(\frac{101.8}{100})^{(y-1994)}$

$log_{10}2=(y-1994)log_{10}(\frac{101.8}{100})$

$y-1994=\frac{log_{10}2}{log_{10}1.018}=38.8537$

$y$ ≈ $2033$

∴ By 2033, the population doubles.

4 0

3 years ago

Joshua finds the perimeter of the following composite figure composed of a square and right triangle so that a braid may be cut

Bumek [7]

The perimeter of a shape is the sum of its side lengths.

<em>Joshua miscalculated the third length of the triangle</em>

First, we calculate the third length (x) of the triangle.

Because the triangle is right-angled, we can make use of Pythagoras.

So we have:

$x^2 = 7^2 + 7^2$

$x^2 = 2(7^2)$

Take square roots of both sides

$x = \sqrt{2(7^2)$

$x = 7\sqrt{2$

The other shape in the figure is a square.

So, the perimeter (P) is

$P = 7 + x + x + x + 7$ ---- i.e. the sum of the visible lengths

So, we have:

$P = 7 + 7\sqrt2 + 7\sqrt2 + 7\sqrt2+ 7$

Evaluate like terms:

$P = 14 + 21\sqrt2$

Hence, Joshua's error is that:

<em>He miscalculated the third length of the triangle</em>