All categories

3 years ago

13. Attend to precision. Justify each step in the solution of 5x + 15 = 0 below by stating a property

Mathematics

1 answer:

tigry1 [53]3 years ago

8 0

Answer:

55/7

Step-by-step explanation:

si:|fvdegregregregrerg4grgrfewgreggtrgr343g

You might be interested in

ΔHAT is similar to ΔCAN.

LenKa [72]

In similar triangles, the ratios of corresponding sides are equal.

This means

y/6=5/(4+5)

Cross multiply

y=5*6/(4+5)=30/9=10/3 (or 3 1/3)

6 0

3 years ago

Read 2 more answers

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

Vsevolod [243]

Answer:

<em>18</em> 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:

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

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

So,as shown above <em>18 values for n are possible.</em>

3 0

3 years ago

What is the equation of the quadratic graph with a focus of (3, 1) and a directrix of y = 5?

telo118 [61]

Answer:

$f(x)=-\frac{1}{8}(x-3)^2+3$

Step-by-step explanation:

We want to find the equation of the parabola with a focus of $(3,1)$ and directrix $y=5$ .

Considering the directrix, the quadratic graph must open downwards.

The equation of this parabola is given by the formula,

$(x-h)^2=4p(y-k)$ , where $(h,k)$ is the vertex of the parabola.

The axis of this parabola meets the directrix at $(3,5)$ .

Since the vertex is the midpoint of the focus and the point of intersection of the axis of the parabola and the directrix,

$h=\frac{3+3}{2}=3$ and $k=\frac{5+1}{2}=3$ .

The equation of the parabola now becomes,

$(x-3)^2=4p(y-3)$ .

Also $|p|$ is the distance between the vertex and the directrix.

$|p|=2$

This implies that $p=-2\:or\:2$ .

Since the parabola turns downwards,

$p=-2$ .

Our equation now becomes,

$(x-3)^2=4(-2)(y-3)$ .

$(x-3)^2=-8(y-3)$ .

We make y the subject to get,

$y=-\frac{1}{8}(x-3)^2+3)$ .

This is the same as

$f(x)=-\frac{1}{8}(x-3)^2+3)$ .

4 0

2 years ago

Can anyone help me please? I have no idea what to do! :'(

liberstina [14]

Answer:

7) 10^(3/2)

8) 2^(1/6)

9) 2^(5/4)

10) 5^(5/4)

Step-by-step explanation:

7) (√10)^3 = 10^(3/2)

8) 6 root 2 = 2^(1/6)

9) (4 root 2)^5 = 2^(5/4)

10) (4 root 5)^5 = 5^(5/4)

5 0

3 years ago

Find the equation in standard form of the line with slope

Serhud [2]

$\bf \begin{array}{lllll}
&x_1&y_1\\
% (a,b)
&({{ 5}}\quad ,&{{ 7}})
\end{array}
\\\\\\
% slope = m
slope = {{ m}}= \cfrac{rise}{run} \implies \cfrac{7}{2}
\\\\\\
% point-slope intercept
\stackrel{\textit{point-slope form}}{y-{{ y_1}}={{ m}}(x-{{ x_1}})}\implies y-7=\cfrac{7}{2}(x-5)\implies y-7=\cfrac{7}{2}x-\cfrac{35}{2}
\\\\\\
y=\cfrac{7}{2}x-\cfrac{35}{2}+7\implies \stackrel{standard~form}{-\cfrac{7}{2}x+y=-\cfrac{21}{2}}$

3 0

3 years ago