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3 years ago

9

You have $59 in a bank account. You need at least $200 to avoid bank fees. Wrote and solve and inequality to find how much more

money you should deposit.

1 answer:

makvit [3.9K]3 years ago

5 0

X represents the amount of money needed.
x<span>≥ 200</span>

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Which statement is it please! let me know !!

garik1379 [7]

The first one i think

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2 years ago

At the end of 8 years, Anna paid $400 interest for a loan. If the simple annual interest rate was 4%, how much did Anna borrow?

antiseptic1488 [7]

$1250

Use the formula i = prt

In this case, we don't know P or Principal amount

i = 400

r = .04 *found by diving 4/100*

n = 8

Set formula to solve for P like: P= i/r(n)

P=400/.04(8)

P= 400/.32

P = 1250

6 0

3 years ago

Read 2 more answers

The diagrams shows a 3cm x 5cm x 4cm cuboid.

Vera_Pavlovna [14]

Answer:

a) 5.83 cm

b) 34.45°

Step-by-step explanation:

a) From Pythagoras theorem of right triangles, given right triangle ABC:

AB² + BC² = AC²

Therefore:

AC² = 5² + 3²

AC² = 25 + 9 = 34

AC = √34

AC = 5.83 cm

b) From triangle ACD, AC = 5.83 cm, AD = 4 cm and ∠A = 90°.

From Pythagoras theorem of right triangles, given right triangle ACD:

AD² + AC² = DC²

Therefore:

DC² = 5.83² + 4²

DC² = 34 + 16 = 50

DC = √50

DC = 7.07 cm

Let ∠ACD be x. Therefore using sine rule:

$\frac{4}{sin(x)}=\frac{7.07}{sin(90)} \\ sin(x)=\frac{4*sin(90))}{7.07}=0.5657\\ x=sin^{-1}(0.5657)\\x=34.45^o$

6 0

2 years ago

Directions: Calculate the area of a circle using 3.14x the radius

Leokris [45]

$\qquad\qquad\huge\underline{{\sf Answer}}♨$

As we know ~

Area of the circle is :

$\qquad \sf \dashrightarrow \:\pi {r}^{2}$

And radius (r) = diameter (d) ÷ 2

[ radius of the circle = half the measure of diameter ]

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<h3>Problem 1</h3>

$\qquad \sf \dashrightarrow \:r = d \div 2$

$\qquad \sf \dashrightarrow \:r = 4.4\div 2$

$\qquad \sf \dashrightarrow \:r = 2.2 \: mm$

Now find the Area ~

$\qquad \sf \dashrightarrow \: \pi {r}^{2}$

$\qquad \sf \dashrightarrow \:3.14 \times {(2.2)}^{2}$

$\qquad \sf \dashrightarrow \:3.14 \times {4.84}^{}$

$\qquad \sf \dashrightarrow \:area \approx 15.2 \: \: mm {}^{2}$

・．━━━━━━━†━━━━━━━━━．・

<h3>problem 2</h3>

$\qquad \sf \dashrightarrow \:r = d \div 2$

$\qquad \sf \dashrightarrow \:r = 3.7 \div 2$

$\qquad \sf \dashrightarrow \:r = 1.85 \: \: cm$

Bow, calculate the Area ~

$\qquad \sf \dashrightarrow \: \pi {r}^{2}$

$\qquad \sf \dashrightarrow \:3.14 \times (1.85) {}^{2}$

$\qquad \sf \dashrightarrow \:3.14 \times 3.4225 {}^{}$

$\qquad \sf \dashrightarrow \:area \approx 10.75 \: \: cm {}^{2}$

・．━━━━━━━†━━━━━━━━━．・

<h3>Problem 3 </h3>

$\qquad \sf \dashrightarrow \:\pi {r}^{2}$

$\qquad \sf \dashrightarrow \:3.14 \times (8.3) {}^{2}$

$\qquad \sf \dashrightarrow \:3.14 \times 68.89$

$\qquad \sf \dashrightarrow \:area \approx216.31 \: \: cm {}^{2}$

・．━━━━━━━†━━━━━━━━━．・

<h3>Problem 4</h3>

$\qquad \sf \dashrightarrow \:r = d \div 2$

$\qquad \sf \dashrightarrow \:r = 5.8 \div 2$

$\qquad \sf \dashrightarrow \:r = 2.9 \: \: yd$

now, let's calculate area ~

$\qquad \sf \dashrightarrow \:3.14 \times {(2.9)}^{2}$

$\qquad \sf \dashrightarrow \:3.14 \times 8.41$

$\qquad \sf \dashrightarrow \:area \approx26.41 \: \: yd {}^{2}$

・．━━━━━━━†━━━━━━━━━．・

<h3>problem 5</h3>

$\qquad \sf \dashrightarrow \:r = d \div 2$

$\qquad \sf \dashrightarrow \:r = 1 \div 2$

$\qquad \sf \dashrightarrow \:r = 0.5 \: \: yd$

Now, let's calculate area ~

$\qquad \sf \dashrightarrow \:\pi {r}^{2}$

$\qquad \sf \dashrightarrow \:3.14 \times (0.5) {}^{2}$

$\qquad \sf \dashrightarrow \:3.14 \times 0.25$

$\qquad \sf \dashrightarrow \:area \approx0.785 \: \: yd {}^{2}$

・．━━━━━━━†━━━━━━━━━．・

<h3>problem 6</h3>

$\qquad \sf \dashrightarrow \:\pi {r}^{2}$

$\qquad \sf \dashrightarrow \:3.14 \times {(8)}^{2}$

$\qquad \sf \dashrightarrow \:3.14 \times 64$

$\qquad \sf \dashrightarrow \:area = 200.96 \: \: yd {}^{2}$

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8 0

2 years ago

At the market, 8 batteries cost $10 .How much do 6 batteries cost

vagabundo [1.1K]

$10 / 8 = 1.25
Soo...
$6 x 1.25 = 7.50

6 0

3 years ago

Read 2 more answers