All categories

3 years ago

Rationalize the denominator and simplify. 5/4-√3

1 answer:

4 0

I assume the denominator is (4-√3)
Multiple both top and bottom by (4+√3)
[5(4+√3)]/[(4-√3)(4+√3)]
=[20+5√3]/[4²-(√3)²]
=[20+5√3]/13

You might be interested in

$25000 at 4.25% for 2 year

lianna [129]

Answer:

Step-by-step

Monthly Payment $1,088.41

Total Interest Paid $1,121.77

Total Paid $26,121.77

Mark brainliest if this answer is correct please

6 0

2 years ago

Find x3 − 2y2 − 3x3 + z4 if x = 3, y = 5, and z = −3.

Oksana_A [137]

3^3-2(5)^2-3(3)^3+-3(4)
27-50-81+-12
-116

4 0

3 years ago

Jacey obtains a 30-year 6/2 ARM at 4% with a 2/6 cap structure in the amount of $224,500. What is the monthly payment during the

Juli2301 [7.4K]

General Idea:

We need to make use of the below formula to find the monthly payment..

$Monthly \; Payment\; =\; \frac{P \times \frac{r}{12}}{(1-(1+\frac{r}{12})^{-m})} \\ \\ Where:\\ P\; is\; Principal\\ r\; is\; rate\; in\; decimal\; form\\ m\; is\; number\; of\; monthly\; payments$

Applying the concept:

Given:

$P=\$224,500\\ r=4\%=0.04\\ m=30\; year \times 12 \; months/year=360\\$

Substituting the given in the formula we will get the monthly payment.

$Monthly\; Payment\; =\; \frac{224500 \times \frac{0.04}{12}}{(1-(1+\frac{0.04}{12})^{-360})} =\frac{\frac{8980}{12}}{(1-0.301796)} =\frac{748.3333}{0.698204} \\ \\ Monthly \; Payment= \$1071.7975$

Conclusion:

The monthly payment during the initial period is <u>$1072</u>.

3 0

3 years ago

Read 2 more answers

Mary owned n sheep and Sam had exactly 4 times as many sheep as Mary.

Murljashka [212]

Answer:

least value of n is 9

Step-by-step explanation:

Mary has n sheep. Then she buys 17 extra sheeps.

Mray = n + 17 sheeps

Sam had exactly 4 times as mary. Then Sam sold 8 of his sheep.

Sam = 4n - 8 sheep.

Sam has more sheep than Mary

Sam sheep is greater than Mary

4n - 8 > n + 17

Collect like terms

4n - n > 17 + 8

3n > 25

Divide both sides by 3

n > 25/3

n > 8.3333

Note: the value of n must be whole

Therefore, least value of n is 9

8 0

3 years ago

A sausage company makes two different kinds of hot dogs, regular and all-beef. Each pound of all-beef hot dogs requires 0.75 lb

Gre4nikov [31]

Answer:

880 lbs of all-beef hot dogs
2000 lbs of regular hot dogs
maximum profit is $3320

Step-by-step explanation:

We can let x and y represent the number of pounds of all-beef and regular hot dogs produced, respectively. Then the problem constraints are ...

.75x + 0.18y ≤ 1020 . . . . . . limit on beef supply
.30y ≤ 600 . . . . . . . . . . . . . limit on pork supply
.2x + .2y ≥ 500 . . . . . . . . . . limit on spice supply

And the objective is to maximize

p = 1.50x + 1.00y

The graph shows the constraints, and that the profit is maximized at the point (x, y) = (880, 2000).

2000 pounds of regular and 880 pounds of all-beef hot dogs should be produced. The associated maximum profit is $3320.

8 0

3 years ago