Oval (S) and A, B permanent, straight line d

(C) 9<br>(D) 8<br>(B) 10<br>11<br>2. How many times does the digit 0 occur in two-digit numbers?

Charra [1.4K]

Answer:

9 times.

Step-by-step explanation:

The two digit numbers are 10 - 99.

Zero appears 9 times in the group of numbers.

10, 20, 30, 40, 50, 60, 70, 80, 90.

Therefore, the correct option should be C.

Hope this helps.

8 0

3 years ago

Read 2 more answers

If the triangles in the figure are similar, what is the length of side b? A. 36 B. 40 C. 32.4 D. 45

butalik [34]

Since ABC and DEF are similar then we can write:.

CB/FE = CA/FD
9/90 = b/360
b/360 = 1/10
b = 360/10
b = 36

8 0

4 years ago

Read 2 more answers

the student council at seaside highschool is making t-shirts to sell for a fundraiser at a price of 13 dollars a piece the cost

ankoles [38]

Answer:

299

Step-by-step explanation:

5 0

3 years ago

Hello Please help me.

olga55 [171]

Answer:

Step-by-step explanation:

1) D

2) B

3) no solution

y ≤ 0.5x - 4 is area C including the red line

y ≥ 0.5x + 5 is area A including the blue line

3 0

3 years ago

a quantity with an initial value of 5900 decays exponentially at a rate of 0.85% every 7 years. what is the value of the quantit

laila [671]

The value of the quantity after 99 months is 5228.58.

Given that, an initial value = 5900, rate of decay = 0.85% and time period = 7 years.

<h3>What is the exponential decay?</h3>

The quantity decreases slowly after which the rate of change and the rate of growth decreases over a period of time rapidly. This decrease in growth is calculated by using the exponential decay formula.

The exponential decay formula formula is $Amount=Initial\ value(1-\frac{r}{100} )^{\frac{t}{n} }$

Here, the value $=5900(1-\frac{0.85}{100} )^{\frac{99}{7} }$

= 5900× $(0.9915)^{14.14}$

= 5900×0.8862

= 5228.58

Therefore, the value of the quantity after 99 months is 5228.58.

Learn more about the exponential decay here:

brainly.com/question/13674608.

#SPJ1

4 0

1 year ago