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3 years ago

A grocery store sells a bag of 7 oranges for $2.80. What is the unit cost?

Mathematics

2 answers:

Elanso [62]3 years ago

6 0

Answer:

The quadratic equations and their solutions are;

9 ± √33 /4 = 2x² - 9x + 6.

4 ± √6 /2 = 2x² - 8x + 5.

9 ± √89 /4 = 2x² - 9x - 1.

4 ± √22 /2 = 2x² - 8x - 3.

Explanation:

Any quadratic equation of the form, ax² + bx + c = 0 can be solved using the formula x = -b ± √b² - 4ac / 2a. Here a, b, and c are the coefficients of the x², x, and the numeric term respectively.

We have to solve all of the five equations to be able to match the equations with their solutions.

2x² - 8x + 5, here a = 2, b = -8, c = 5. x = -b ± √b² - 4ac / 2a = -(-8) ± √(-8)² - 4(2)(5) / 2(2) = 8 ± √64 - 40/4. 24 can also be written as 4 × 6 and √4 = 2. So x = 8 ± 2√6 / 2×2= 4±√6/2.

2x² - 10x + 3, here a = 2, b = -10, c = 3. x =-b ± √b² - 4ac / 2a =-(-10) ± √(-10)² - 4(2)(3) / 2(4) = 10 ± √100 + 24/4. 124 can also be written as 4 × 31 and √4 = 2. So x = 10 ± 2√31 / 2×2 = 5 ± √31 /2.

2x² - 8x - 3, here a = 2, b = -8, c = -3. x = -b ± √b² - 4ac / 2a = -(-8) ± √(-8)² - 4(2)(-3) / 2(2) = 8 ± √64 + 24/4. 88 can also be written as 4 × 22 and √4 = 2. So x = 8 ± 2√22 / 2×2 = 4± √22/2.

2x² - 9x - 1, here a = 2, b = -9, c = -1. x = -b ± √b² - 4ac / 2a = -(-9) ± √(-9)² - 4(2)(-1) / 2(2) = 9 ± √81 + 8/4. x = 9 ± √89 / 4.

2x² - 9x + 6, here a = 2, b = -9, c = 6. x = -b ± √b² - 4ac / 2a = -(-9) ± √(-9)² - 4(2)(6) / 2(2) = 9 ± √81 - 48/4. x = 9 ± √33 / 4 .

mark me as brilinist please

SIZIF [17.4K]3 years ago

6 0

Answer:

Step-by-step explanation:

2.80/7=.4

7 x 0.4 = 2.8

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What is the equation of the line that has a slope of -4 and goes through (-1,6)?

OLga [1]

Answer:

y = - 4x + 2

Step-by-step explanation:

The equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

Here m = - 4, thus

y = - 4x + c ← is the partial equation

To find c substitute (- 1, 6) into the partial equation

6 = 4 + c ⇒ c = 6 - 4 = 2

y = - 4x + 2 ← equation of line

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4 years ago

One angle of an isosceles triangle measures 90°. Which other angles could be in that isosceles triangle?

Alika [10]

Hello from MrBillDoesMath!

Answer:

45 (all three: 45, 45, 90)

Discussion:

Let "x" be the base angle in the isosceles triangle. As base angles are equal in an isosceles triangle and a triangle contains 180 degrees:

x + x + 90 = 180 =>

2x + 90 = 180 => subtract 90 from both sides

2x = 180 -90 =>

2x = 90 => divide both sides by 2

x = 45

Thank you,

MrB

3 0

3 years ago

Read 2 more answers

Your annual clothing bill went up from $677.15 last year to $783.17 this year. What is your annual percent of increase from last

ss7ja [257]

Answer:

15.7% just got it right

Step-by-step explanation:

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3 years ago

(x^2y+e^x)dx-x^2dy=0

klio [65]

It looks like the differential equation is

$\left(x^2y + e^x\right) \,\mathrm dx - x^2\,\mathrm dy = 0$

Check for exactness:

$\dfrac{\partial\left(x^2y+e^x\right)}{\partial y} = x^2 \\\\ \dfrac{\partial\left(-x^2\right)}{\partial x} = -2x$

As is, the DE is not exact, so let's try to find an integrating factor µ(x, y) such that

$\mu\left(x^2y + e^x\right) \,\mathrm dx - \mu x^2\,\mathrm dy = 0$

*is* exact. If this modified DE is exact, then

$\dfrac{\partial\left(\mu\left(x^2y+e^x\right)\right)}{\partial y} = \dfrac{\partial\left(-\mu x^2\right)}{\partial x}$

We have

$\dfrac{\partial\left(\mu\left(x^2y+e^x\right)\right)}{\partial y} = \left(x^2y+e^x\right)\dfrac{\partial\mu}{\partial y} + x^2\mu \\\\ \dfrac{\partial\left(-\mu x^2\right)}{\partial x} = -x^2\dfrac{\partial\mu}{\partial x} - 2x\mu \\\\ \implies \left(x^2y+e^x\right)\dfrac{\partial\mu}{\partial y} + x^2\mu = -x^2\dfrac{\partial\mu}{\partial x} - 2x\mu$

Notice that if we let µ(x, y) = µ(x) be independent of y, then ∂µ/∂y = 0 and we can solve for µ :

$x^2\mu = -x^2\dfrac{\mathrm d\mu}{\mathrm dx} - 2x\mu \\\\ (x^2+2x)\mu = -x^2\dfrac{\mathrm d\mu}{\mathrm dx} \\\\ \dfrac{\mathrm d\mu}{\mu} = -\dfrac{x^2+2x}{x^2}\,\mathrm dx \\\\ \dfrac{\mathrm d\mu}{\mu} = \left(-1-\dfrac2x\right)\,\mathrm dx \\\\ \implies \ln|\mu| = -x - 2\ln|x| \\\\ \implies \mu = e^{-x-2\ln|x|} = \dfrac{e^{-x}}{x^2}$