Rewrite each sum as a product. Then identify the number of terms in each expression

5 3/4÷5/9<br>what's answer

anastassius [24]

The answer to the math problem is 10.35

5 0

2 years ago

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$5 PAYPAL + BRAINLIEST TO ANSWER THS

expeople1 [14]

Answer:

Quadratic Equation:

$3x^2=2x +5$

$\text{Standard Form: } 3x^2-2x-5=0$

From the standard form of a Quadratic Function, we get:

$a=3,\:b=-2,\:c=-5$

Discriminant:

$\Delta=\left(-2\right)^2-4\cdot \:3\left(-5\right)$

$\Delta=\left(-2\right)^2+4\cdot \:3\cdot \:5$

$\Delta=64$

From the discriminant, we conclude that the equation will have two real solutions.

State that:

$b^2-4ac$

$b^2-4ac =0:\text{The equation has 1 real solution}$

$b^2-4ac >0:\text{The equation has 2 real solutions}$

By the way, solving the equation given:

$x=\frac{2\pm\sqrt{64}}{2\cdot \:3}$

$x=\frac{2\pm\sqrt{64}}{6}$

$x=\frac{2\pm8}{6}$

$x_{1} =\frac{10}{6}=\frac{5}{3}$

$x_{2}=\frac{-6}{6} =-1$

5 0

3 years ago

(-4)^2 <br> (-3)^3<br> Evaluate

iris [78.8K]

(-4)^2= 16

(-3)^3= -27

7 0

3 years ago

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If anyone is good at grade 8 math problems pls help answer these questions thanks!! :DD

wolverine [178]

Answer:

53 Meters.

Step-by-step explanation:

34*34 + 41*41 = 2837.

1156 + 1681 = 2837.

The square root of 2837 would be your answer.

34*34 + 41*41 = 53.26349 * 53.26349

a*a + b*b = c*c.

a² + b² = c².

5 0

1 year ago

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An aeroplane X whose average speed is 50°km/hr leaves kano airport at 7.00am and travels for 2 hours on a bearing 050°. It then

Zigmanuir [339]

Answer:

(a)123 km/hr

(b)39 degrees

Step-by-step explanation:

Plane X with an average speed of 50km/hr travels for 2 hours from P (Kano Airport) to point Q in the diagram.

Distance = Speed X Time

Therefore: PQ =50km/hr X 2 hr =100 km

It moves from Point Q at 9.00 am and arrives at the airstrip A by 11.30am.

Distance, QA=50km/hr X 2.5 hr =125 km

Using alternate angles in the diagram:

$\angle Q=110^\circ$

(a)First, we calculate the distance traveled, PA by plane Y.

Using Cosine rule

$q^2=p^2+a^2-2pa\cos Q\\q^2=100^2+125^2-2(100)(125)\cos 110^\circ\\q^2=34175.50\\q=184.87$ km$

SInce aeroplane Y leaves kano airport at 10.00am and arrives at 11.30am

Time taken =1.5 hour

Therefore:

Average Speed of Y

$=184.87 \div 1.5\\=123.25$ km/hr\\\approx 123$ km/hr (correct to three significant figures)$

(b)Flight Direction of Y

Using Law of Sines

$\dfrac{p}{\sin P} =\dfrac{q}{\sin Q}\\\dfrac{125}{\sin P} =\dfrac{184.87}{\sin 110}\\123 \times \sin P=125 \times \sin 110\\\sin P=(125 \times \sin 110) \div 184.87\\P=\arcsin [(125 \times \sin 110) \div 184.87]\\P=39^\circ $ (to the nearest degree)$

The direction of flight Y to the nearest degree is 39 degrees.

7 0

3 years ago