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3 years ago

$800 at 3.8% interest for 6 months What is the interest? What is the new balance?

Mathematics

2 answers:

Furkat [3]3 years ago

4 0

Use the equation I/PT = R I is the interest P is the money before investment and T is the time

viva [34]3 years ago

4 0

Multiply then divide

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An office building rents for $40,000 the first year, then increases by 0.5% each year. Use summation notation to represent the t

Sholpan [36]

Given the initial rent is $40000, and the rent is increased by 0.5% each year, the rent in the third year will be given by:
FV=P(1+r)^n
FV=40000(1.05)^n
Thus the answer for the sum of payments for 3 years will be:
The summation is 1.05 raised to n years times $40000, where the years is from 0 to 3.

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4 years ago

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Peggy completed 5/6 of the math homework and Al completed 4/5 of the math homework. Did Peggy or Al complete more of the math ho

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Peggy completed more because more equal parts that are smaller with more shaded means it has to be more.

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3 years ago

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Solve the following expression: -3+10 - -4 • -5+-6

serg [7]

Answer:

-19

Step-by-step explanation:

-3+10- (-4) × (-5)+(-6)

-3+10+4 × (-5)-6

4×(-5) = -20 -3+10-20-6 = -19

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3 years ago

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Drag the tiles to the correct boxes to complete the pairs not all tiles will be used match each quadratic graph to its respectiv

ch4aika [34]

Answer:

Part 1) The function of the First graph is $f(x)=(x-3)(x+1)$

Part 2) The function of the Second graph is $f(x)=-2(x-1)(x+3)$

Part 3) The function of the Third graph is $f(x)=0.5(x-6)(x+2)$

See the attached figure

Step-by-step explanation:

we know that

The quadratic equation in factored form is equal to

$f(x)=a(x-c)(x-d)$

where

a is the leading coefficient

c and d are the roots or zeros of the function

Part 1) First graph

we know that

The solutions or zeros of the first graph are

x=-1 and x=3

The parabola open up, so the leading coefficient a is positive

The function is equal to

$f(x)=a(x-3)(x+1)$

Find the value of the coefficient a

The vertex is equal to the point (1,-4)

substitute and solve for a

$-4=a(1-3)(1+1)$

$-4=a(-2)(2)$

$a=1$

therefore

The function is equal to

$f(x)=(x-3)(x+1)$

Part 2) Second graph

we know that

The solutions or zeros of the first graph are

x=-3 and x=1

The parabola open down, so the leading coefficient a is negative

The function is equal to

$f(x)=a(x-1)(x+3)$

Find the value of the coefficient a

The vertex is equal to the point (-1,8)

substitute and solve for a

$8=a(-1-1)(-1+3)$

$8=a(-2)(2)$

$a=-2$

therefore

The function is equal to

$f(x)=-2(x-1)(x+3)$

Part 3) Third graph

we know that

The solutions or zeros of the first graph are

x=-2 and x=6

The parabola open up, so the leading coefficient a is positive

The function is equal to

$f(x)=a(x-6)(x+2)$

Find the value of the coefficient a

The vertex is equal to the point (2,-8)

substitute and solve for a

$-8=a(2-6)(2+2)$

$-8=a(-4)(4)$

$a=0.5$

therefore

The function is equal to

$f(x)=0.5(x-6)(x+2)$

3 0

4 years ago

What is the slope between the points (-2,4) and (-3,5)

defon

Answer:

$y = \frac{1}{-1}x+2$

3 0

3 years ago