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2 years ago

Find the type and number of solution for g(x) = x2 -14x = -50

Mathematics

1 answer:

inysia [295]2 years ago

3 0

Answer:

Step-by-step explanation:

x² -14x+50= 0

this equation has 2 solutions because is a quadratic

the solurions are imaginary roots because the discriminant is less then 0

b²-4ac = (-14)²-4*1*50 = 194-200= -6

to find the actual roots use the quadratic formula

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Round 1,129.872 To The Nearest Tenth

aleksklad [387]

1,129.872

The 8 is in the tenth place. There is a 7 in the hundredths place. Since 7 is greater than or equal to 5, we round the 8 up. So it becomes a 9.

1129.9

Answer: 1129.9

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3 years ago

To the nearest tenth, what is the area of a circle with circumference 35.6 centimeters?

AnnyKZ [126]

The answer is A. 100.9 cm2

7 0

3 years ago

Read 2 more answers

A garden is shaped in the form of a regular heptagon (seven-sided), MNSRQPO. A circle with center T and radius 25m circumscribes

Alenkinab [10]

The relationship between the sides MN, MS, and MQ in the given regular heptagon is $\dfrac{1}{MN} = \dfrac{1}{MS} + \dfrac{1}{MQ}$

The area to be planted with flowers is approximately <u>923.558 m²</u>

The reason the above value is correct is as follows;

The known parameters of the garden are;

The radius of the circle that circumscribes the heptagon, r = 25 m

The area left for the children playground = ΔMSQ

Required;

The area of the garden planted with flowers

Solution:

The area of an heptagon, is;

$A = \dfrac{7}{4} \cdot a^2 \cdot cot \left (\dfrac{180 ^{\circ}}{7} \right )$

The interior angle of an heptagon = 128.571°

The length of a side, S, is given as follows;

$\dfrac{s}{sin(180 - 128.571)} = \dfrac{25}{sin \left(\dfrac{128.571}{2} \right)}$

$s = \dfrac{25}{sin \left(\dfrac{128.571}{2} \right)} \times sin(180 - 128.571) \approx 21.69$

$The \ apothem \ a = 25 \times sin \left ( \dfrac{128.571}{2} \right) \approx 22.52$

The area of the heptagon MNSRQPO is therefore;

$A = \dfrac{7}{4} \times 22.52^2 \times cot \left (\dfrac{180 ^{\circ}}{7} \right ) \approx 1,842.94$

$MS = \sqrt{(21.69^2 + 21.69^2 - 2 \times 21.69 \times21.69\times cos(128.571^{\circ})) \approx 43.08$

By sine rule, we have

$\dfrac{21.69}{sin(\angle NSM)} = \dfrac{43.08}{sin(128.571 ^{\circ})}$

$sin(\angle NSM) =\dfrac{21.69}{43.08} \times sin(128.571 ^{\circ})$

$\angle NSM = arcsin \left(\dfrac{21.69}{43.08} \times sin(128.571 ^{\circ}) \right) \approx 23.18^{\circ}$

∠MSQ = 128.571 - 2*23.18 = 82.211

The area of triangle, MSQ, is given as follows;

$Area \ of \Delta MSQ = \dfrac{1}{2} \times 43.08^2 \times sin(82.211^{\circ}) \approx 919.382^{\circ}$

The area of the of the garden plated with flowers, $A_{req}$ , is given as follows;

$A_{req}$ = Area of heptagon MNSRQPO - Area of triangle ΔMSQ

Therefore;

$A_{req}$ = 1,842.94 - 919.382 ≈ 923.558

The area of the of the garden plated with flowers, $A_{req}$ ≈ <u>923.558 m²</u>

Learn more about figures circumscribed by a circle here:

brainly.com/question/16478185

6 0

3 years ago

Given: -1/4x > 4. Choose the solution set.

vampirchik [111]

Answer:

x < -16, none of the solution sets are correct for the given equation.

Step-by-step explanation:

-1/4x > 4, multiply both sides by -4, don't forget when you multiply or divide by a negative you have to flip the sign.

x < -16

none of the solution sets are correct for the given equation.

5 0

3 years ago

If a certain cannon is fired from a height of 8.8 meters above the ground, at a certain angle, the height of the cannonball ab

Dennis_Churaev [7]

Answer:

It would take approximately 6.50 second for the cannonball to strike the ground.

Step-by-step explanation:

Consider the provided function.

$h(t)=-4.9t^2+30.5t+8.8$

We need to find the time takes for the cannonball to strike the ground.

Substitute h(t) = 0 in above function.

$-4.9t^2+30.5t+8.8=0$

Multiply both sides by 10.

$-49t^2+305t+88=0$

For a quadratic equation of the form $ax^2+bx+c=0$ the solutions are: $x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Substitute a = -49, b = 305 and c=88

$t=\frac{-305+\sqrt{305^2-4\left(-49\right)88}}{2\left(-49\right)}=-\frac{-305+\sqrt{110273}}{98}\\t = \frac{-305-\sqrt{305^2-4\left(-49\right)88}}{2\left(-49\right)}= \frac{305+\sqrt{110273}}{98}$

Ignore the negative value of t as time can't be a negative number.

Thus,

$t=\frac{305+\sqrt{110273}}{98}\approx6.50$

Hence, it would take approximately 6.50 second for the cannonball to strike the ground.

6 0

3 years ago