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3 years ago

Select the two values of x that are roots of this equation 2x^2-11x+15=0

1 answer:

3 0

X=5/2 and 3. Factor the equation find numbers that will sum up too -11 involving the coefficient of 2, and that multiply into 15. 2x too the power of 2 -11x +15=0 (2x-5)(x-3) =0

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Write the equation in slope intercept form when slope = 6/5 and the y- intercept =-8

Ratling [72]

Step-by-step explanation:

Exercícios:

9. Sobre o estudo de equações, escreva V para as afirmações verdadeiras e F para as falsas.

a) ( ) A expressão 17 - 4x + 25 representa uma equação.

b) ( ) O número 23 representa um dos membros da equação -9x + 4 = 23.

c) ( ) Na equação x + 4 = 3x + 20, a incógnita é x.

d) ( ) A equação 3y - 15 = 0 não tem segundo membro.

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3 years ago

Which is the most appropriate unit to measure the length of a slice of pizza?

Ne4ueva [31]

A centimetre is the most appropriate unit

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3 years ago

Line segment ABCD is split into unequal segments by points A, B, C, and D. The length of segment AD is 28 cm, and the distance b

Oksana_A [137]

Answer:

4 cm

Step-by-step explanation:

The midpoint segment has length ...

... L = AB/2 + BC + CD/2 = 16 cm

The entire segment has length ...

... AD = AB + BC + CD = 28 cm

If we subtract AD from 2L, we have ...

... 2L - AD - 2L = 2(AB/2 +BC + CD/2) - (AB +BC +CD) = 2(16 cm) -28 cm = 4 cm

... AB +2BC + CD -AB -BC -CD = 4 cm . . . . remove parentheses

... BC = 4 cm . . . . simplify

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3 years ago

Part 1. Determine the molar mass of a 0.622-gram sample of gas having a volume of 2.4 L at 287 K and 0.850 atm. Show your work.

sdas [7]

If a sample of gas is a 0.622-gram, volume of 2.4 L at 287 K and 0.850 atm. Then the molar mass of the gas is 7.18.

<h3>What is an ideal gas equation?</h3>

The ideal gas equation is given below.

$PV = nRT$

The equation can be written as

$\rm PV = \dfrac{m}{M} \ RT\\\\\\MPV = mRT$

Where M is the molar mass, P is the pressure, V is the volume, R is the universal gas constant, T is the temperature, and m is the mass of the gas.

Then the molar mass of a 0.622-gram sample of gas has a volume of 2.4 L at 287 K and 0.850 atm.

V = 2.4 L = 0.0024

P = 0.85 atm = 86126.25 Pa

T = 287 K

m = 0.622

R = 8.314

Then we have

$\begin{aligned} \rm M \times 86126.25 \times 0.0024&= 0.622 \times 8.314 \times 287\\\\\rm M &= 7.18 \end{aligned}$