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2 years ago

What is the solution of the quadratic equation. Equation: 2x + 7x - 5= 10

Mathematics

1 answer:

Slav-nsk [51]2 years ago

4 0

Step-by-step explanation:

2x+7x-5=10

2x+7x= 10+5= 15

2x ×9=15

2x =15-9=5

x=5/2=2.5

<h3>hope it helped</h3>

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A juggler tosses a ball into the air . The balls height, h and time t seconds can be represented by the equation h(t)= -16t^2+40

malfutka [58]

PART A

The given equation is

$h(t) = - 16 {t}^{2} + 40t + 4$

In order to find the maximum height, we write the function in the vertex form.

We factor -16 out of the first two terms to get,

$h(t) = - 16 ({t}^{2} - \frac{5}{2} t) + 4$

We add and subtract

$- 16(- \frac{5}{4} )^{2}$

to get,

$h(t) = - 16 ({t}^{2} - \frac{5}{2} t) + - 16( - \frac{5}{4})^{2} - -16( - \frac{5}{4})^{2} + 4$

We again factor -16 out of the first two terms to get,

$h(t) = - 16 ({t}^{2} - \frac{5}{2} t + ( - \frac{5}{4})^{2} ) - -16( - \frac{5}{4})^{2} + 4$

This implies that,

$h(t) = - 16 ({t}^{2} - \frac{5}{2} t + ( - \frac{5}{4}) ^{2} ) + 16( \frac{25}{16}) + 4$

The quadratic trinomial above is a perfect square.

$h(t) = - 16 ( t- \frac{5}{4}) ^{2} +25+ 4$

This finally simplifies to,

$h(t) = - 16 ( t- \frac{5}{4}) ^{2} +29$

The vertex of this function is

$V( \frac{5}{4} ,29)$

The y-value of the vertex is the maximum value.

Therefore the maximum value is,

$29$

PART B

When the ball hits the ground,

$h(t) = 0$

This implies that,

$- 16 ( t- \frac{5}{4}) ^{2} +29 = 0$

We add -29 to both sides to get,

$- 16 ( t- \frac{5}{4}) ^{2} = - 29$

This implies that,

$( t- \frac{5}{4}) ^{2} = \frac{29}{16}$

$t- \frac{5}{4} = \pm \sqrt{ \frac{29}{16} }$

$t = \frac{5}{4} \pm \frac{ \sqrt{29} }{4}$

$t = \frac{ 5 + \sqrt{29} }{4} = 2.60$

or

$t = \frac{ 5 - \sqrt{29} }{4} = - 0.10$

Since time cannot be negative, we discard the negative value and pick,

$t = 2.60s$

8 0

3 years ago

We considered the differences between the temperature readings in January 1 of 1968 and 2008 at 51 locations in the continental

mr Goodwill [35]

Answer:

$1.1-2.02\frac{4.9}{\sqrt{50}}=-0.30$

$1.1+2.02\frac{4.9}{\sqrt{50}}=2.50$

So on this case the 90% confidence interval would be given by (-0.30;2.50)

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X=1.1$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

s=4.9 represent the sample standard deviation

n=51 represent the sample size

Solution to the problem

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=51-1=50$

Since the Confidence is 0.90 or 90%, the value of $\alpha=0.1$ and $\alpha/2 =0.05$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.05,50)".And we see that $t_{\alpha/2}=2.02$

Now we have everything in order to replace into formula (1):

$1.1-2.02\frac{4.9}{\sqrt{50}}=-0.30$

$1.1+2.02\frac{4.9}{\sqrt{50}}=2.50$

So on this case the 90% confidence interval would be given by (-0.30;2.50)

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3 years ago

Which of the following describes the end behavior of y = -x^2+bx+c as x approaches either positive or negative infinity?

Pavlova-9 [17]

The end behavior model of y is -x^2.
As x approaches either positive or neg infinity, -x^2 will always be negative.
Thus, "B) y approaches neg infinity" is the correct one.
Hope that will help :))

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3 years ago

Which statements are true? Check all that apply

xxTIMURxx [149]

Answer:

A,C,D

Step-by-step explanation:

Snap

Jona_Fl16

7 0

3 years ago

Given: BD bisects ABC. Auxiliary EA is drawn such that AE || BD. Auxiliary BE is an extension of BC

AnnZ [28]

The required proof is given in the table below:

$\begin{tabular}{|p{4cm}|p{6cm}|} 
 Statement & Reason \\ [1ex] 
1. $\overline{BD}$ bisects $\angle ABC$ & 1. Given \\
2. \angle DBC\cong\angle ABD & 2. De(finition of angle bisector \\ 
3. $\overline{AE}$||$\overline{BD}$ & 3. Given \\ 
4. \angle AEB\cong\angle DBC & 4. Corresponding angles \\
5. \angle AEB\cong\angle ABD & 5. Transitive property of equality \\ 
6. \angle ABD\cong\angle BAE & 6. Alternate angles
\end{tabular}$
$\begin{tabular}{|p{4cm}|p{6cm}|}
7. \angle AEB\cong\angle BAE & 7. Transitive property of equality \\
8. \overline{EB}\cong\overline{AB} & 8. From 7. $\Delta ABE$ is isosceles \\
9. EB = AB & 9. De(finition of congruence \\ 10. $\frac{AD}{DC}=\frac{EB}{BC}$ & 10. Triangle proportionality theorem \\
11. $\frac{AD}{DC}=\frac{AB}{BC}$ & 11. Substitution Property of equality \\[1ex] 
\end{tabular}
$

7 0

3 years ago

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