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3 years ago

7

Determine the type and number of solutions of −4x2 − 3x + 7 = 0

1 answer:

Mrrafil [7]3 years ago

5 0

Since 3^2 - (-4)(7) must be bigger than 0

Then it has two real roots.

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Solve (2x + 3) - (9x - 3)

Diano4ka-milaya [45]

Answer:

-7x + 6

Step-by-step explanation:

3 0

2 years ago

City A had a population of 10000 in the year 1990. City A’s population grows at a constant rate of 3% per year. City B has a pop

Georgia [21]

Answer:

City A and city B will have equal population 25years after 1990

Step-by-step explanation:

Given

Let

$t \to$ years after 1990

$A_t \to$ population function of city A

$B_t \to$ population function of city B

<u>City A</u>

$A_0 = 10000$ ---- initial population (1990)

$r_A =3\%$ --- rate

<u>City B</u>

$B_{10} = \frac{1}{2} * A_{10}$ ----- t = 10 in 2000

$A_{20} = B_{20} * (1 + 20\%)$ ---- t = 20 in 2010

Required

When they will have the same population

Both functions follow exponential function.

So, we have:

$A_t = A_0 * (1 + r_A)^t$

$B_t = B_0 * (1 + r_B)^t$

Calculate the population of city A in 2000 (t = 10)

$A_t = A_0 * (1 + r_A)^t$

$A_{10} = 10000 * (1 + 3\%)^{10}$

$A_{10} = 10000 * (1 + 0.03)^{10}$

$A_{10} = 10000 * (1.03)^{10}$

$A_{10} = 13439.16$

Calculate the population of city A in 2010 (t = 20)

$A_t = A_0 * (1 + r_A)^t$

$A_{20} = 10000 * (1 + 3\%)^{20}$

$A_{20} = 10000 * (1 + 0.03)^{20}$

$A_{20} = 10000 * (1.03)^{20}$

$A_{20} = 18061.11$

From the question, we have:

$B_{10} = \frac{1}{2} * A_{10}$ and $A_{20} = B_{20} * (1 + 20\%)$

$B_{10} = \frac{1}{2} * A_{10}$

$B_{10} = \frac{1}{2} * 13439.16$

$B_{10} = 6719.58$

$A_{20} = B_{20} * (1 + 20\%)$

$18061.11 = B_{20} * (1 + 20\%)$

$18061.11 = B_{20} * (1 + 0.20)$

$18061.11 = B_{20} * (1.20)$

Solve for B20

$B_{20} = \frac{18061.11}{1.20}$

$B_{20} = 15050.93$

$B_{10} = 6719.58$ and $B_{20} = 15050.93$ can be used to determine the function of city B

$B_t = B_0 * (1 + r_B)^t$

For: $B_{10} = 6719.58$

We have:

$B_{10} = B_0 * (1 + r_B)^{10}$

$B_0 * (1 + r_B)^{10} = 6719.58$

For: $B_{20} = 15050.93$

We have:

$B_{20} = B_0 * (1 + r_B)^{20}$

$B_0 * (1 + r_B)^{20} = 15050.93$

Divide $B_0 * (1 + r_B)^{20} = 15050.93$ by $B_0 * (1 + r_B)^{10} = 6719.58$

$\frac{B_0 * (1 + r_B)^{20}}{B_0 * (1 + r_B)^{10}} = \frac{15050.93}{6719.58}$

$\frac{(1 + r_B)^{20}}{(1 + r_B)^{10}} = 2.2399$

Apply law of indices

$(1 + r_B)^{20-10} = 2.2399$

$(1 + r_B)^{10} = 2.2399$ --- (1)

Take 10th root of both sides

$1 + r_B = \sqrt[10]{2.2399}$

$1 + r_B = 1.08$

Subtract 1 from both sides

$r_B = 0.08$

To calculate $B_0$ , we have:

$B_0 * (1 + r_B)^{10} = 6719.58$

Recall that: $(1 + r_B)^{10} = 2.2399$

So:

$B_0 * 2.2399 = 6719.58$

$B_0 = \frac{6719.58}{2.2399}$

$B_0 = 3000$

Hence:

$B_t = B_0 * (1 + r_B)^t$

$B_t = 3000 * (1 + 0.08)^t$

$B_t = 3000 * (1.08)^t$

The question requires that we solve for t when:

$A_t = B_t$

Where:

$A_t = A_0 * (1 + r_A)^t$

$A_t = 10000 * (1 + 3\%)^t$

$A_t = 10000 * (1 + 0.03)^t$

$A_t = 10000 * (1.03)^t$

and

$B_t = 3000 * (1.08)^t$

$A_t = B_t$ becomes

$10000 * (1.03)^t = 3000 * (1.08)^t$

Divide both sides by 10000

$(1.03)^t = 0.3 * (1.08)^t$

Divide both sides by $(1.08)^t$

$(\frac{1.03}{1.08})^t = 0.3$

$(0.9537)^t = 0.3$

Take natural logarithm of both sides

$\ln(0.9537)^t = \ln(0.3)$

Rewrite as:

$t\cdot\ln(0.9537) = \ln(0.3)$

Solve for t

$t = \frac{\ln(0.3)}{ln(0.9537)}$

$t = 25.397$

Approximate

$t = 25$

7 0

2 years ago

If you get this you are a critical thinker.

Alchen [17]

Answer:

5

Step-by-step explanation:

If you killed 30 and there were 34 people there not including me then there would be 5 left.

4 0

2 years ago

Read 2 more answers

Solve for x: 2|х - 3 |+1 = 7

Rasek [7]

When using problems like this use mathaway love

5 0

2 years ago

2 u + 12 < 23<br> 0,1 d + 8 > 0<br> 3 - 4 r > 7<br> 13 < 2 z + 3<br> 6 ≥ 9 - 0, 15 i

BartSMP [9]

Answer:

Part 1) $u< 5.5$

Part 2) $d > -80$

Part 3) $r < -1$

Part 4) $z>5$

Part 5) $i\geq 20$

Step-by-step explanation:

<u><em>The question is</em></u>

Solve each inequality for the indicated variable

Part 1) we have

$2u+12$

subtract 12 both sides

$2u< 23-12\\2u$

Divide by 2 both sides

$u< 5.5$

The solution is the interval (-∞,5.5)

In a number line the solution is the shaded area at left of u=5.5 (open circle)

The number 5.5 is not included in the solution

Part 2) we have

$0.1d+8 >0$

subtract 8 both sides

$0.1d > -8$

Divide by 0.1 both sides

$d > -80$

The solution is the interval (-80,∞)

In a number line the solution is the shaded area at right of d=-80 (open circle)

The number -80 is not included in the solution

Part 3) we have

$3-4r> 7$

Subtract 3 both sides

$-4r>7-3\\-4r>4$

Divide by -4 both sides

Remember that, when you multiply or divide both sides of an inequality by a negative number, you must reverse the inequality symbol

$r < -1$

The solution is the interval (-∞,-1)

In a number line the solution is the shaded area at left of r=-1.1 (open circle)

The number -1.1 is not included in the solution

Part 4) we have

$13< 2z+3$

Subtract 3 both sides

$13-3$

Divide by 2 both sides

$5$

Rewrite

$z>5$

The solution is the interval (5,∞)

In a number line the solution is the shaded area at right of z=5 (open circle)

The number 5 is not included in the solution

Part 5) we have

$6\geq 9-0.15i$

Subtract 9 both sides

$6-9\geq -0.15i\\-3\geq -0.15i$

Divide by -0.15 both sides

Remember that, when you multiply or divide both sides of an inequality by a negative number, you must reverse the inequality symbol

$20\leq i$

Rewrite

$i\geq 20$

The solution is the interval [20,∞)

In a number line the solution is the shaded area at right of i=20 (closed circle)

The number 20 is included in the solution

6 0

2 years ago