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2 years ago

If a = and b = -1 what is the value of ab -b^2

Mathematics

1 answer:

svet-max [94.6K]2 years ago

8 0

Answer:

-2

Step-by-step explanation:

To solve you need to replace the variables with their numbers.

3(-1) - (-1^2)

-1 x -1 = -1

3 x (-1) = -3

-3 - (-1) =

-2

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3 years ago

1.What is the area of this parallelogram?

dexar [7]

Question 1: Option D

Area of the parallelogram = 132 cm²

Question 2: Option B

Area of the parallelogram = 3.60 ft²

Solution:

Question 1:

Base of the parallelogram = 6 + 5 = 11 cm

Height of the parallelogram = 12 cm

Area of the parallelogram = Base × Height

= 11 × 12

Area of the parallelogram = 132 cm²

Option D is the correct answer.

Question 2:

Base of the parallelogram = 1.9 + 0.5 = 2.4 ft

Height of the parallelogram = 1.5 ft

Area of the parallelogram = Base × Height

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Area of the parallelogram = 3.60 ft²

Option B is the correct answer.

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3 years ago

Is -5 greater than -10

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Answer:

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3 years ago

A quantity with an initial value of 6200 decays continuously at a rate of 5.5% per month. What is the value of the quantity afte

ELEN [110]

Answer:

410.32

Step-by-step explanation:

Given that the initial quantity, Q= 6200

Decay rate, r = 5.5% per month

So, the value of quantity after 1 month, $q_1 = Q- r \times Q$

$q_1 = Q(1-r)\cdots(i)$

The value of quantity after 2 months, $q_2 = q_1- r \times q_1$

$q_2 = q_1(1-r)$

From equation (i)

$q_2=Q(1-r)(1-r) \\\\q_2=Q(1-r)^2\cdots(ii)$

The value of quantity after 3 months, $q_3 = q_2- r \times q_2$

$q_3 = q_2(1-r)$

From equation (ii)

$q_3=Q(1-r)^2(1-r)$

$q_3=Q(1-r)^3$

Similarly, the value of quantity after n months,

$q_n= Q(1- r)^n$

As 4 years = 48 months, so puttion n=48 to get the value of quantity after 4 years, we have,

$q_{48}=Q(1-r)^{48}$

Putting Q=6200 and r=5.5%=0.055, we have

$q_{48}=6200(1-0.055)^{48} \\\\q_{48}=410.32$

Hence, the value of quantity after 4 years is 410.32.

4 0

2 years ago

Read 2 more answers