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3 years ago

Factor the following equation2x^2-9x-5

Mathematics

1 answer:

-Dominant- [34]3 years ago

3 0

Answer:

(2x +1 ) (x -5 )

Step-by-step explanation:

2x^2-9x-5

(2x ) (x )

The factors of 5 are 1 and 5

One will be positive and one is negative, but which one

We want to get to -9 in the middle

2* -5 = -10 +1 = -9

(2x +1 ) (x -5 )

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2 years ago

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2 years ago

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2 years ago

find the area of the trapezium whose parallel sides are 25 cm and 13 cm The Other sides of a Trapezium are 15 cm and 15 CM

Snezhnost [94]

$\huge\underline{\red{A}\blue{n}\pink{s}\purple{w}\orange{e}\green{r} -}$

Given - A trapezium ABCD with non parallel sides of measure 15 cm each ! along , the parallel sides are of measure 13 cm and 25 cm

To find - Area of trapezium

Refer the figure attached ~

In the given figure ,

AB = 25 cm

BC = AD = 15 cm

CD = 13 cm

Construction -

$draw \: CE \: \parallel \: AD \: \\ and \: CD \: \perp \: AE$

Now , we can clearly see that AECD is a parallelogram !

$\therefore$ AE = CD = 13 cm

Now ,

$AB = AE + BE \\\implies \: BE =AB - AE \\ \implies \: BE = 25 - 13 \\ \implies \: BE = 12 \: cm$

Now , In ∆ BCE ,

$semi \: perimeter \: (s) = \frac{15 + 15 + 12}{2} \\ \\ \implies \: s = \frac{42}{2} = 21 \: cm$

Now , by Heron's formula

$area \: of \: \triangle \: BCE = \sqrt{s(s - a)(s - b)(s - c)} \\ \implies \sqrt{21(21 - 15)(21 - 15)(21 - 12)} \\ \implies \: 21 \times 6 \times 6 \times 9 \\ \implies \: 12 \sqrt{21} \: cm {}^{2}$

Also ,

$area \: of \: \triangle \: = \frac{1}{2} \times base \times height \\ \\\implies 18 \sqrt{21} = \: \frac{1}{\cancel2} \times \cancel12 \times height \\ \\ \implies \: 18 \sqrt{21} = 6 \times height \\ \\ \implies \: height = \frac{\cancel{18} \sqrt{21} }{ \cancel 6} \\ \\ \implies \: height = 3 \sqrt{21} \: cm {}^{2}$

Since we've obtained the height now , we can easily find out the area of trapezium !

$Area \: of \: trapezium = \frac{1}{2} \times(sum \: of \:parallel \: sides) \times height \\ \\ \implies \: \frac{1}{2} \times (25 + 13) \times 3 \sqrt{21} \\ \\ \implies \: \frac{1}{\cancel2} \times \cancel{38 }\times 3 \sqrt{21} \\ \\ \implies \: 19 \times 3 \sqrt{21} \: cm {}^{2} \\ \\ \implies \: 57 \sqrt{21} \: cm {}^{2}$

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1 year ago

Two horses are ready to return to their barn after a long workout session at the track. The horses are at coordinates H(1,10) an

lana [24]

Distance between points $\left(x_1,y_1\right)$ and $\left( x_2,y_2 \right)$ is

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2$

Distance from H to B:

$[tex]d=\sqrt{(10-(-3))^2+(1-(-9))^2}=\sqrt{169+100}=\sqrt{269}$ d=\sqrt{(1-(-3))^2+(10-(-9))^2}=\sqrt{16+361}=\sqrt{376}[/tex] units.

Distance from Z to B:

$d=\sqrt{(10-(-3))^2+(1-(-9))^2}=\sqrt{169+100}=\sqrt{269}$ units.

Horse Z is closer to the barn.

(The conversion to meters is not required; the question does not ask for actual distances, so "units" is OK.)

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3 years ago

Factor the following equation2x^2-9x-5 ​

Factor the following equation2x^2-9x-5