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3 years ago

Given: ABCD is a rhombus. Prove: △AEB is a right triangle.

Mathematics

1 answer:

lukranit [14]3 years ago

4 0

Answer:

Step-by-step explanation:

∵ ABCD is a rhombus

∴ AC is ⊥ on BD

∴ ΔAEB is right angled triangle

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Scott invests $1000 at a bank that offers 6% compounded annually.Write an equation to model the growth of the investment

gladu [14]

Answer:

Step-by-step explanation:

Using the formula for the growth of investment:

.....[1]

where,

A is the amount after t year

P is the Principal

r is the growth rate in decimal

As per the statement:

Scott invests $1000 at a bank that offers 6% compounded annually.

⇒P = $1000 and r = 6% = 0.06

substitute these in [1] we get;

⇒

Therefore, an equation to model the growth of the investment is,

7 0

2 years ago

The table shows the number of markers in different numbers of boxes.

frutty [35]

Answer:

Explanation:

4 0

3 years ago

Read 2 more answers

Write two decimals greater than luke's decimal and less than bekka's decimal

Mashcka [7]

.23 and .24 is your answer.

8 0

3 years ago

Write an equation of the line in slope-intercept form that passes through the point (3, 8) with slope 4?

iren [92.7K]

Slope intercept form:
Y = mx + b
m = slope, b = y intercept
Given the slope of 4
Y = 4x + b
Plug in the point (3,8)
8 = 4(3) + b
8 = 12 + b, b = -4
Y intercept = -4

Solution: y = 4x - 4

8 0

3 years ago

What is the probability of rolling a four then a three

spayn [35]

Answer:

$\frac{1}{36}$

Step-by-step explanation:

Total possibilities when we a roll a die at a time are 6

given we should have four for first time and then three

let us assume we rolled the dice we may get 1,2,3,4,5,6(any of these) the probability to get 4 is

PROBABILITY= $\frac{\textrm{FAVOURABLE CHANCES}}{\textrm{TOTAL CHANCES}}$

Favourable chances=1

Total chances=6

Probability= $\frac{1}{6}$

the prabability to get 4 in first roll is $\frac{1}{6}$ .

let us assume we rolled the dice for second time again we may get 1,2,3,4,5,6(any of these) the probability to get 3 is

Favourable chances=1

total chances=6

probability= $\frac{1}{6}$

the probability to get 3 in second roll irrespective of first one is $\frac{1}{6}$

the probability to get 4 in first time and then 3 is

The probability to occur both events at a time is multiplication of individual probabilities

So,

probablility to get 4 in first roll= $\frac{1}{6}$

probability to get 3 in second roll= $\frac{1}{6}$

probability to occur both at a same time is = $\frac{1}{6}$ $\times$ $\frac{1}{6}$ = $\frac{1}{36}$

6 0

3 years ago