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2 years ago

Is the bisector of ∠ABC and is the bisector of ∠ACB. Also,

Mathematics

1 answer:

drek231 [11]2 years ago

8 0

Answer: No, it is not the bisector of angle ACB

Angle ABC has the vertex at point B. The middle point determines the vertex. This is why angle ABC is the exact same as angle CBA. Each time, B is in the middle.

However, angle ACB has point C in the middle so we're talking about a completely different angle. This is not a bisector because the line doesn't even go through point C at all. The line would need to curve, but bisectors must be straight lines.

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What is 6-2p=p+9 ?Because I don't know it

nikklg [1K]

Hey there!

Firstly, you have to $\bold{simplify}$ each of the sides of your equation
$\bold{6-2p-p+9\downarrow}$
$\bold{6+(-2p)=p+9\downarrow}$
$\rightarrow\bold{-2p+6=p+9}$
Now that you have that portion out of the way, we could move on to our second step... which is to $\bold{subtract}$ by the value of $\bold{p}$ on each of your sides
$\bold{-2p+6-p-p+9-p\downarrow}$
$\rightarrow\bold{-3p+6-9}$ (you get this because we solve the particular equation we added above)
Thirdly, you have to $\bold{subtract}$ by the number $\bold{6}$ on each of your sides
$\bold{-3p+6-6=9-6\downarrow}$
$\bold{Cancel:6-6}$ because it equals to $\bold{0}$
$\bold{Keep:9-6}$ because it brings us closer and closer to the answer of $\bold{p}$
Fourthly, we have to $\bold{divide}$ by the number $\bold{-3}$ on each of your sides
$\bold{\frac{-3p}{-3}=\frac{3}{-3}}$
$\boxed{\boxed{\bold{Answer:p=-1}}}$ $\checkmark$

Good luck on your assignment and enjoy your day!

~ $\frak{LoveYourselfFirst:)}$

8 0

3 years ago

Adam has 16.45 kg of flour, he uses 6.4 kg to make hot cross buns. The remaining flour is exactly enough to make 15 batches of s

Doss [256]

Answer:

Each batch of scones needed 0.67 kg of flour

Step-by-step explanation:

16.45 - 6.4 = 10.05kg (remaining)

15 batches = 10.05kg

1 batch = 0.67kg

6 0

3 years ago

Keegan and Annie are working on an experiment in the science lab. Their mixture contains ingredients that weigh 5.22 grams, 5.7

Elza [17]

Answer:

Step-by-step explanation:

5 0

2 years ago

Read the story.

Rudiy27

Answer:

20% i am not 100% sure. if i am wrong, i am sorry

3 0

2 years ago

The state education commission wants to estimate the fraction of tenth grade students that have reading skills at or below the e

Viefleur [7K]

Answer:

The 95% confidence interval for the population proportion of tenth graders reading at or below the eighth grade level is (0.2087, 0.2507).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

For this problem, we have that:

Suppose a sample of 1537 tenth graders is drawn. Of the students sampled, 1184 read above the eighth grade level. So 1537 - 1184 = 353 read at or below this level. Then

$n = 1537, \pi = \frac{353}{1537} = 0.2297$

95% confidence level

So $\alpha = 0.05$ , z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$ , so $Z = 1.96$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2297 - 1.96\sqrt{\frac{0.2297*0.7703}{1537}} = 0.2087$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2297 + 1.96\sqrt{\frac{0.2297*0.7703}{1537}} = 0.2507$

The 95% confidence interval for the population proportion of tenth graders reading at or below the eighth grade level is (0.2087, 0.2507).

7 0

2 years ago