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3 years ago

Please help me with this

Mathematics

1 answer:

goldenfox [79]3 years ago

7 0

For part A, if P is the center of that triangle, then PR and PT have the same length; therefore, triangle RPT is isosceles. For part B, by the definition of an incenter...if P is an incenter, then it is the place where all the angle bisectors meet. Therefore, angles SRP and PRT are congruent, as are angles STP and PTR. Since the vertex angle measures 64, then each of the base angles by the isosceles triangle theorem measure 58. Half of 58 makes the base angles within the smaller triangle measure 29. And if both of those measure 29, by the triangle angle-sum theorem, 180-29-29 = 122 And that's the measure of angle RPT. Eek.

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Brady studied 1.5 hours on Monday, 0.75 hours on Tuesday, 1.25 hours on Wednesday, and 1 hour on Thursday. How

Advocard [28]

Brady studied for 4.5 hours that week

How we determine the total study hours?

The total study hours for Brady is the sum of the hours used in studying on Monday, Tuesday, Wednesday and Thursday.

He studied for 1.5 hours on Monday

He studied for 0.75 hours on Tuesday

He studied again for 1.25 hours on Wednesday

Lastly, he studied for 1 hour on Thursday, the last day

Total number of study hours=1.5+0.75+1.25+1

Total number of study hours=4.50

Find out more about study hour on: brainly.com/question/876733

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2 years ago

Let Q(x, y) be the predicate "If x < y then x2 < y2," with domain for both x and y being R, the set of all real numbers.

Levart [38]

Answer:

a) Q(-2,1) is false

b) Q(-5,2) is false

c)Q(3,8) is true

d)Q(9,10) is true

Step-by-step explanation:

Given data is $Q(x,y)$ is predicate that $x$ then $x^{2}$ . where $x,y$ are rational numbers.

when $x=-2, y=1$

Here $-2$ that is $x$ satisfied. Then

$(-2)^{2}$

$4$ this is wrong. since $4>1$

That is $x^{2}$ $>y^{2}$ Thus $Q(x,y)$ $=Q(-2,1)$ is false.

Assume $Q(x,y)=Q(-5,2)$ .

That is $x=-5, y=2$

Here $-5$ that is $x$ this condition is satisfied.

Then

$(-5)^{2}$

$25$ this is not true. since $25>4$ .

This is similar to the truth value of part (a).

Since in both $x$ satisfied and $x^{2} >y^{2}$ for both the points.

if $Q(x,y)=Q(3,8)$ that is $x=3$ and $y=8$

Here $3$ this satisfies the condition $x$ .

Then $3^{2}$

$9$ This also satisfies the condition $x^{2}$ .

Hence $Q(3,8)$ exists and it is true.

Assume $Q(x,y)=Q(9,10)$

Here $9$ satisfies the condition $x$

Then $9^{2}$

$81$ satisfies the condition $x^{2}$ .

Thus, $Q(9,10)$ point exists and it is true. This satisfies the same values as in part (c)

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3 years ago

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Juli2301 [7.4K]

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3 years ago

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nasty-shy [4]

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morpeh [17]

Answer:

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Step-by-step explanation:

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3 years ago

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