Triangle ABC is isosceles.

A number minus 9 is twelve

TiliK225 [7]

Answer:

The answer is 21

Step-by-step explanation:

All you have to do is 9+12

Brainliest??

5 0

3 years ago

Read 2 more answers

A car purchased for $24,000 is expected to lose value, or depreciate, at a rate of 8% per year. This situation can be modeled by

algol [13]

$\bf \qquad \textit{Amount for Exponential Decay} \\\\ A=P(1 - r)^t\qquad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{initial amount}\dotfill &24000\\ r=rate\to 8\%\to \frac{8}{100}\dotfill &0.08\\ t=\textit{elapsed time}\dotfill &t\\ \end{cases} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{current value}}{\stackrel{A}{\stackrel{\downarrow }{P(t)}}}=24000(1-0.08)^{\stackrel{\stackrel{\textit{years}}{\downarrow }}{t}}$

5 0

2 years ago

A middle school has the fifth and sixth grades. There are 100 fifth-grade boys and 110 fifth-grade girls. There are 7 fewer sixt

Mama L [17]

The ratio of girls to boys in the middle school = 213 : 193

Number of fifth-grade boys = 100

Number of fifth grade girls = 110

There are 7 fewer sixth-grade boys than fifth-grade boys

Number of sixth-grade boys = 100 - 7

Number of sixth-grade boys = 93

There are 10 more sixth-grade girls than sixth-grade boys

Number of sixth-grade girls = 93 + 10

Number of sixth-grade girls = 103

Total number of girls = Number of sixth-grade girls + Number of fifth-grade girls

Total number of girls = `103 + 110

Total number of girls = 213

Total number of boys = Number of sixth-grade boys + Number of fifth-grade boys

Total number of boys = `93 + 100

Total number of boys = 193

The ratio of girls to boys in the middle school = 213 : 193

Learn more on ratio here: brainly.com/question/25927869

4 0

2 years ago

Solve rational inequality x²+x-6/x²-3x-4≤0?

Fittoniya [83]

Answer:

Step-by-step explanation:

x

2

+

x

−

6

=

(

x

+

3

)

(

x

−

2

)

x

2

−

3

x

−

4

=

(

x

−

4

)

(

x

+

1

)

Each of the linear factors occurs precisely once, so the sign of the given rational expression will change at each of the points where one of the linear factors is zero. That is at:

x

=

−

3

,

−

1

,

2

,

4

Note that when

x

is large, the

x

2

terms will dominate the values of the numerator and denominator, making both positive.

Hence the sign of the value of the rational expression in each of the intervals

(

−

∞

,

−

3

)

,

(

−

3

,

−

1

)

,

(

−

1

,

2

)

,

(

2

,

4

)

and

(

4

,

∞

)

follows the pattern

+

−

+

−

+

. Hence the intervals

(

−

3

,

−

1

)

and

(

2

,

4

)

are both part of the solution set.

When

x

=

−

1

or

x

=

4

, the denominator is zero so the rational expression is undefined. Since the numerator is non-zero at those values, the function will have vertical asymptotes at those points (and not satisfy the inequality).

When

x

=

−

3

or

x

=

2

, the numerator is zero and the denominator is non-zero. So the function will be zero and satisfy the inequality at those points.

Hence the solution is:

x

∈

[

−

3

,

−

1

)

∪

[

2

,

4

)

graph{(x^2+x-6)/(x^2-3x-4) [-10, 10, -5, 5]}

8 0

2 years ago

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Ninety-one percent of products come off the line within product specifications. Your quality control department selects 15 produ

Allisa [31]

Answer:

Probability of stopping the machine when $X < 9$ is 0.0002

Probability of stopping the machine when $X < 10$ is 0.0013

Probability of stopping the machine when $X < 11$ is 0.0082

Probability of stopping the machine when $X < 12$ is 0.0399

Step-by-step explanation:

There is a random binomial variable $X$ that represents the number of units come off the line within product specifications in a review of $n$ Bernoulli-type trials with probability of success $0.91$ . Therefore, the model is ${15 \choose x} (0.91) ^ {x} (0.09) ^ {(15-x)}$ . So:

$P (X < 9) = 1 - P (X \geq 9) = 1 - [{15 \choose 9} (0.91)^{9}(0.09)^{6}+...+{ 15 \choose 15}(0.91)^{15}(0.09)^{0}] = 0.0002$

$P (X < 10) = 1 - P (X \geq 10) = 1 - [{15 \choose 10}(0.91)^{10}(0.09)^{5}+...+{15 \choose 15} (0.91)^{15}(0.09)^{0}] = 0.0013$

$P (X < 11) = 1 - P (X \geq 11) = 1 - [{15 \choose 11}(0.91)^{11}(0.09)^{4}+...+{15 \choose 15} (0.91)^{15}(0.09)^{0}] = 0.0082$

$P (X < 12) = 1- P (X \geq 12) = 1 - [{15 \choose 12}(0.91)^{12}(0.09)^{3}+...+{15 \choose 15} (0.91)^{15}(0.09)^{0}] = 0.0399$

Probability of stopping the machine when $X < 9$ is 0.0002

Probability of stopping the machine when $X < 10$ is 0.0013

Probability of stopping the machine when $X < 11$ is 0.0082

Probability of stopping the machine when $X < 12$ is 0.0399

8 0

2 years ago