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3 years ago

Point K is the midpoint of ¹segment JLJK = 2x+4JL=12

Mathematics

1 answer:

Eddi Din [679]3 years ago

8 0

Step-by-step explanation:

$2(2x + 4) = 12 \\ 4x + 8 = 12 \\ 4x = 4 \\ x = 1$

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lilavasa [31]

Answer:

SB or ? = 9

Step-by-step explanation:

Since the two triangles are similar.

((18 + 6) / 6) = 24 / 6 = 4

[Scale factor between ∆BCU and ∆SUT]

SB or ? = 12 – 12/4 = 12 – 3 = 9

4 0

2 years ago

Read 2 more answers

How do you determine where f(x)=cos^(-1)(lnx) is continuous?

olga nikolaevna [1]

$\ln x$ is continuous over its domain, all real $x>0$ .

Meanwhile, $\cos^{-1}y$ is defined for real $-1\le y\le1$ .

If $y=\ln x$ , then we have $-1\le \ln x\le1\implies \dfrac1e\le x\le e$ as the domain of $\cos^{-1}(\ln x)$ .

We know that if $f$ and $g$ are continuous functions, then so is the composite function $f\circ g$ .

Both $\cos^{-1}y$ and $\ln x$ are continuous on their domains (excluding the endpoints in the case of $\cos^{-1}y$ ), which means $\cos^{-1}(\ln x)$ is continuous over $\dfrac1e$ .

7 0

3 years ago

Suppose the roots of the polynomial $x^2 - mx + n$ are positive prime integers (not necessarily distinct). Given that $m < 20

Vsevolod [243]

Answer:

18 values for n are possible.

Step-by-step explanation:

Given the quadratic polynomial:

$x^2 - mx + n$

such that:

Roots are positive prime integers and

$m < 20$

To find:

How many possible values of $n$ are there ?

Solution:

First of all, let us have a look at the sum and product of a quadratic equation.

If the quadratic equation is:

$Ax^{2} +Bx+C$

and the roots are: $\alpha$ and $\beta$

Then sum of roots, $\alpha+\beta = -\frac{B}{A}$

Product of roots, $\alpha \beta = \frac{C}{A}$

Comparing the given equation with standard equation, we get:

A = 1, B = -m and C = n

Sum of roots, $\alpha+\beta = -\frac{-m}{1} = m$

Product of roots, $\alpha \beta = \frac{n}{1} = n$

We are given that $m$

$\alpha$ and $\beta$ are positive prime integers such that their sum is less than 20.

Let us have a look at some of the positive prime integers:

2, 3, 5, 7, 11, 13, 17, 23, 29, .....

Now, we have to choose two such prime integers from above list such that their sum is less than 20 and the roots can be repetitive as well.

So, possible combinations and possible value of $n (= \alpha \times \beta)$ are:

$1.\ 2, 2\Rightarrow n = 2\times 2 = 4\\2.\ 2, 3 \Rightarrow n = 6\\3.\ 2, 5 \Rightarrow n = 10\\4.\ 2, 7\Rightarrow n = 14\\5.\ 2, 11 \Rightarrow n = 22\\6.\ 2, 13 \Rightarrow n = 26\\7.\ 2, 17 \Rightarrow n = 34\\8.\ 3, 3\Rightarrow n = 3\times 3 = 9\\9.\ 3, 5 \Rightarrow n = 15\\10.\ 3, 7 \Rightarrow n = 21\\$

$11.\ 3, 11\Rightarrow n = 33\\12.\ 3, 13 \Rightarrow n = 39\\13.\ 5, 5 \Rightarrow n = 25\\14.\ 5, 7 \Rightarrow n = 35\\15.\ 5, 11 \Rightarrow n = 55\\16.\ 5, 13 \Rightarrow n = 65\\17.\ 7, 7 \Rightarrow n = 49\\18.\ 7, 11 \Rightarrow n = 77$