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3 years ago

15

Lisa loaned John $5000 at an interest rate of 6%. He repaid her $5750 to cover the principal and interest. How long did he borro

w the money?

1 answer:

horrorfan [7]3 years ago

4 0

Answer:

time in years = 2.5 years

Step-by-step explanation:

time = interest / (principal amount x rate)

time = 750 / 5000 x 0.06

time = 750 / 300

time = 2.5 years

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Angle G in trapezoid FGHJ measures 135 degrees. Jasmine reflects the trapezoid over the x-axis. What is the measure of the image

butalik [34]

Answer:

Triangle EFG has vertices E(-3, 4), F(-5, -1), and G(1, 1)

Step-by-step explanation:

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3 years ago

Read 2 more answers

Will mark the Brainliest !!!!! A team of 5 boys and 4 girls will be chosen from a group of 16 boys and 13 girls,

vfiekz [6]

Answer:

Step-by-step explanation:

The total number of permutations of boys and girls on the team are:

¹⁶P₅*¹³P₄

1.

Bob will be one of the fixed boys to be picked. Hence, actually 4 boys are to be picked from 15. The permutations of girls being picked remains the same.

Probability = Permutations with Bob as one of the boys / Total permutations

Probability = (¹⁵P₄*¹³P₄) / (¹⁶P₅*¹³P₄) = ¹⁵P₄ / ¹⁶P₅

Probability = $\frac{15!}{(15-4)!}/\frac{16!}{(16-5)!}$ = 15! / 16! = 1/16

2.

Now, Bob is one of the fixed boys and Jane is one of the fixed girls. Hence, actually 4 boys are to be picked from 15 and 3 girls are to be picked from 12.

Probability = Permutations with bob as one of the boys and jane as one of the girls / Total permutations

Probability = (¹⁵P₄*¹²P₃) / (¹⁶P₅*¹³P₄) = (1/16)*(1/13) = 1/208

3.

Now, the probability that at least Jane or Bob will be picked has been asked. This probability is a combination of three probabilities:

Probability = (Probability that only Bob will be picked) + (Probability that only Jane will be picked) + (Probability that both will be picked)

Probability = 1/16 + 1/13 + 1/208 = 0.123

4.

Total teams possible = ¹⁶P₅*¹³P₄ = 8994585600 teams are possible

5 0

3 years ago

Please help! I've been working on this for a few days and I just don't understand, it's due in a few hours. Thank you.

never [62]

Answer:

Part A: α = arc tan (y/x) = tan⁻¹ (y/x)

Part B: quadrant II α = arc tan (- y/x) = arc (- tan (y/x)) = - tan⁻¹ (y/x) 180°>α>90°

quadrant III α = arc tan (-y)/(-x) = arc tan (y/x) = tan⁻¹ (y/x) 270°>α>180°

quadrant IV α = arc tan (- y/x) = arc (- tan (y/x)) = - tan⁻¹ (y/x) 360°>α>270°

Part C: quadrant II 180°>α>90° α = tan⁻¹ (-6/1) = - tan⁻¹ 6 = 180° - 80.53° = 99.47°

Step-by-step explanation:

PART A:

In this case we will use trigonometric function tanα to calculate angle α:

tanα = y/x => α = arc tan (y/x) = tan⁻¹ (y/x)

This formula is use in general way and in first quadrant 90°>α>0°

Part B:

But in the other quadrants you must know to use unit circle to reduce angle from II, III and IV quadrant to the first quadrant.

If angle is in the quadrant II 180°>α>90° then

tanα = - tan (180°-α)

If angle is in the quadrant III 270°>α>180° then

tanα = tan (α-180°)

If angle is in the quadrant IV 360°>α>270° then

tanα =- tan (360°-α)

Part C:

vector w (x,y) = (-1,6) this vector lies in quadrant II 180°>α>90°

180°- α = tan⁻¹ (-6/1) = - tan⁻¹ 6 = 80.53° => 180° - α = 80.53°

α = 180° - 80.53 = 99.47°

It's not easy to understand this, but it's not easy for me to explain.

God with you!!!

5 0

3 years ago

Detmermine the best method to solve the following equation, then solve the equation. (3x-5)^2=-125

liq [111]

For the given equation;

$(3x-5)^2=-125$

We shall begin by expanding the parenthesis on the left side, after which we would combine all terms on and move all of them to the left side, which shall yield a quadratic equation. Then we shall solve.

Let us begin by expanding the parenthesis;

$\begin{gathered} (3x-5)^2\Rightarrow(3x-5)(3x-5) \\ (3x-5)(3x-5)=9x^2-15x-15x+25 \\ (3x-5)^2=9x^2-30x+25 \end{gathered}$

Now that we have expanded the left side of the equation, we would have;

$\begin{gathered} 9x^2-30x+25=-125 \\ \text{Add 125 to both sides and we'll have;} \\ 9x^2-30x+25+125=-125+125 \\ 9x^2-30x+150=0 \end{gathered}$

We shall now solve the resulting quadratic equation using the quadratic formula as follows;

$\begin{gathered} x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ \text{Where;} \\ a=9,b=-30,c=150 \\ x=\frac{-(-30)\pm\sqrt[]{(-30)^2-4(9)(150)}}{2(9)} \\ x=\frac{30\pm\sqrt[]{900-5400}}{18} \\ x=\frac{30\pm\sqrt[]{-4500}}{18} \\ x=\frac{30\pm\sqrt[]{-900\times5}}{18} \\ x=\frac{30\pm\sqrt[]{-900}\times\sqrt[]{5}}{18} \\ x=\frac{30\pm30i\sqrt[]{5}}{18} \\ \text{Therefore;} \\ x=\frac{30+30i\sqrt[]{5}}{18},x=\frac{30-30i\sqrt[]{5}}{18} \\ \text{Divide all through by 6, and we'll have;} \\ x=\frac{5+5i\sqrt[]{5}}{3},x=\frac{5-5i\sqrt[]{5}}{3} \end{gathered}$

ANSWER:

$x=\frac{5+5i\sqrt[]{5}}{3},x=\frac{5-5i\sqrt[]{5}}{3}$

3 0

1 year ago

Craig left his house at noon and drove 50 miles per hour until 3PM. Then he drove the next 5 hours at 70 miles per hour. Calcula

Amiraneli [1.4K]

Answer:

Step-by-step explanation:

Craig left his house at noon and drove 50 miles per hour until 3PM. Then he drove the next 5 hours at 70 miles per hour. Calculate the average rate of change for the entire trip.

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3 years ago