All categories

1 month ago

Find the x is qs bisects pqr and pqr=82

Mathematics

3 answers:

Svet_ta [12.4K]1 month ago

5 0

Given ∠PQR = 82°, and ray QS bisects it, the ray divides the angle into two equal parts: ∠PQS and ∠RQS, each measuring 82/2 = 41°.

$\bf \stackrel{\measuredangle PQS}{10x+1}=41\implies 10x=40\implies x=\cfrac{40}{10}\implies x=4$

tester [12K]1 month ago

3 0

Answer:

$x=4$

Detailed Solution:

We know that ray QS bisects angle PQR, with ∠PQR measuring 82°. The task is to find the value of x.

Since QS divides ∠PQR equally, both angles created by QS are half of 82°.

$m\angle PQS=\frac{m\angle PQR}{2}$

$10x+1=\frac{82}{2}$

$10x+1=41$

$10x+1-1=41-1$

$10x=40$

$\frac{10x}{10}=\frac{40}{10}$

$x=4$

Hence, the value of x is 4.

Inessa [12.3K]1 month ago

0 0

lnbbgdjdidi

You might be interested in

Shalise competed in a jigsaw puzzle competition where participants are timed on how long they take to complete puzzles of variou

Svet_ta [12483]

Answer:

The small puzzle took 16% of the total time while the large puzzle took 84%.

Step-by-step explanation:

3 0

1 month ago

The dye dilution method is used to measure cardiac output with 3 mg of dye. The dye concentrations, in mg/L, are modeled by c(t)

lawyer [12283]

Answer:

Cardiac output: $F=0.055 L\s$

Step-by-step explanation:

Details: The cardiac output is assessed using the dye dilution technique with 3 mg of dye.

Goal: To determine the cardiac output value.

Solution:

Cardiac output formula: $F=\frac{A}{\int\limits^T_0 {c(t)} \, dt}$ ---1

A = 3 mg

$\int\limits^T_0 {c(t)} \, dt =\int\limits^{10}_0 {20te^{-0.06t}} \, dt$

Utilize integration by parts

$[\int{20te^{-0.6t}} \, dt]^{10}_0=[20t\int{e^{-0.6t} \,dt}-\int[\frac{d[20t]}{dt}\int {e^{-0.6t} \, dt]dt]^{10}_0$

$[\int{20te^{-0.6t}} \, dt]^{10}_0=[\frac{-20te^{-0.6t}}{0.6}+\frac{20}{0.6}\int {e^{-0.6t} \,dt]^{10}_0$

$[\int{20te^{-0.6t}} \, dt]^{10}_0=[\frac{-20te^{-0.6t}}{0.6}+\frac{20e^{-0.6t}}{(0.6)^2}]^{10}_{0}$

$[\int{20te^{-0.6t}} \, dt]^{10}_0=[\frac{-200e^{-6}}{0.6}+\frac{20e^{-6}}{(0.6)^2}]+\frac{20}{(0.60^2}$

$[\int{20te^{-0.6t}} \, dt]^{10}_0=\frac{20(1-e^{-6}}{(0.6)^2}-\frac{200e^{-6}}{0.6}$

$[\int{20te^{-0.6t}} \, dt]^{10}_0\sim {54.49}$

Insert the value into 1

Cardiac output: $F=\frac{3}{54.49}$

Cardiac output: $F=0.055 L\s$

Consequently, Cardiac output: $F=0.055 L\s$

4 0

1 month ago

John's average for making free throws in a basketball game is .80. In a one-and-one free throw situation (where he shoots a seco

babunello [11495]

The likelihood that John scores on the first attempt and misses the second is:
$P(makes\ 1\ basket)=0.8\times(1-0.8)=0.16$

4 0

29 days ago

Read 2 more answers

Keshawn is asked to compare and contrast the domain and range for the two functions.

Zina [12190]

The response is:

Therefore, option 1 is correct.

Hence, the final option is accurate.

Step-by-step breakdown:

Two functions are presented

. The domain represents the x values possible

. The range represents the y values achievable

. Thus, x can take any real number in functions f(x) and g(x)

. So, option 1 is correct.

For the range aspect

$f(x)=5x$ , f(x) can encompass any real number $g(x)=5^x$

. However, for g(x), the range can only include positive numbers

. Consequently, the last option is accurate, while all other options are incorrect.

3 0

6 days ago

Read 2 more answers

Circle D is shown. Line segment F E goes through point D. Line segment C D is shown. Line segment G H goes from one side of the

PIT_PIT [12094]

The answer is Option D: Line G H.

Step-by-step explanation:

Test was conducted on Edge

3 0

18 days ago

Read 2 more answers