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3 years ago

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8,000 is invested in two different accounts earning 3% and 4% interest. If the two accounts earn a total of $295 in interest, ho

w much money was invested at 3%?

2 answers:

scZoUnD [109]3 years ago

8 0

Answer:

The correct answer would be $2,500

Step-by-step explanation:

lesya692 [45]3 years ago

3 0

Answer:

$5,500

Step-by-step explanation:

Let A be the amount invested at 3%, B the amount at 4%.

Total value:

1.A%2BB=8000

Interest:

Multiply eq. 1 by -3 and add to eq. 2,

-3A-3B%2B3A%2B4B=-24000%2B29500

highlight%28B=5500%29

Then from eq. 1,

A%2B5500=8000

highlight%28A=2500%29

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In a naval engagement, one-third of the fleet was captured, one-sixth was sunk, and two ships were destroyed by fire. One-sevent

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Answer:

60 ships.

Step-by-step explanation:

Let the total number of ships in the naval fleet be represented by x

One-third of the fleet was captured = 1/3x

One-sixth was sunk = 1/6x

Two ships were destroyed by fire = 2

Let surviving ships be represented by y

One-seventh of the surviving ships were lost in a storm after the battle = 1/7y

Finally, the twenty-four remaining ships sailed home

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Collect like terms

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Maths functions question!!

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Answer:

5) DE = 7 units and DF = 4 units

6) ST = 8 units

$\textsf{7)} \quad \sf OM=\dfrac{3}{2}\:units$

8) x ≤ -3 and x ≥ 3

Step-by-step explanation:

<u>Information from Parts 1-4:</u>

brainly.com/question/28193969

$f(x)=-x+3$

$g(x)=x^2-9$

A = (3, 0) and C = (-3, 0)

<h3><u>Part (5)</u></h3>

Points A and D are the <u>points of intersection</u> of the two functions.

To find the x-values of the points of intersection, equate the two functions and solve for x:

$\implies g(x)=f(x)$

$\implies x^2-9=-x+3$

$\implies x^2+x-12=0$

$\implies x^2+4x-3x-12=0$

$\implies x(x+4)-3(x+4)=0$

$\implies (x-3)(x+4)=0$

Apply the zero-product property:

$\implies (x-3)= \implies x=3$

$\implies (x+4)=0 \implies x=-4$

From inspection of the graph, we can see that the x-value of point D is <u>negative</u>, therefore the x-value of point D is x = -4.

To find the y-value of point D, substitute the found value of x into one of the functions:

$\implies f(-4)=-(-4)=7$

Therefore, D = (-4, 7).

The length of DE is the difference between the y-value of D and the x-axis:

⇒ DE = 7 units

The length of DF is the difference between the x-value of D and the x-axis:

⇒ DF = 4 units

<h3><u>Part (6)</u></h3>

To find point S, substitute the x-value of point T into function g(x):

$\implies g(4)=(4)^2-9=7$

Therefore, S = (4, 7).

The length ST is the difference between the y-values of points S and T:

$\implies ST=y_S-y_T=7-(-1)=8$

Therefore, ST = 8 units.

<h3><u>Part (7)</u></h3>

The given length of QR (⁴⁵/₄) is the difference between the functions at the same value of x. To find the x-value of points Q and R (and therefore the x-value of point M), subtract g(x) from f(x) and equate to QR, then solve for x:

$\implies f(x)-g(x)=QR$

$\implies -x+3-(x^2-9)=\dfrac{45}{4}$

$\implies -x+3-x^2+9=\dfrac{45}{4}$

$\implies -x^2-x+\dfrac{3}{4}=0$

$\implies -4\left(-x^2-x+\dfrac{3}{4}\right)=-4(0)$

$\implies 4x^2+4x-3=0$

$\implies 4x^2+6x-2x-3=0$

$\implies 2x(2x+3)-1(2x+3)=0$

$\implies (2x-1)(2x+3)=0$

Apply the zero-product property:

$\implies (2x-1)=0 \implies x=\dfrac{1}{2}$

$\implies (2x+3)=0 \implies x=-\dfrac{3}{2}$

As the x-value of points M, Q and P is negative, x = -³/₂.

Length OM is the difference between the x-values of points M and the origin O:

$\implies x_O-x_m=o-(-\frac{3}{2})=\dfrac{3}{2}$

Therefore, OM = ³/₂ units.

<h3><u>Part (8)</u></h3>

The values of x for which g(x) ≥ 0 are the values of x when the parabola is above the x-axis.

Therefore, g(x) ≥ 0 when x ≤ -3 and x ≥ 3.

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