All categories

3 years ago

Plz give me answer anybody fast

Mathematics

1 answer:

vlabodo [156]3 years ago

5 0

Answer:

Step-by-step explanation:

Since AD is an angle bisector then it divides the opposite side into segments that are proportional to the other 2 sides, that is

$\frac{AC}{AB}$ = $\frac{CD}{BD}$ , substitute values

$\frac{AC}{8}$ = $\frac{3}{6}$ ( cross- multiply )

6AC = 24 ( divide both sides by 6 )

AC = 4 cm → A

You might be interested in

A spool of ribbon hold 6.75 m a craft buys 21 spools what is the total cost of the ribbon sells for 2 dollars per meter

kherson [118]

The answer would be $70.875

3 0

3 years ago

please help me with my question due tomorrow morning I will like and mark as brainliest for the first correct answer please

galina1969 [7]

Answer:

p = (2k + 1)i + j

q = 25i + (k + 4)j

p . q = 11

Find the dot product of p and q

(2k + 1)(25) + (k + 4) = 11

50k + 25 + k + 4 = 11

51k = -18

k = - 18/51 = –6/17.✅

b) p.q = |p||q|cos∅

p= (2k + 1)i + j = [2(-6/17) + 1)i + j ]= 5/17i + j

|p| = √ (5/17)² + 1²

|p| = √314/ 17 ( The square root doesn't cover the 17) = 1.0424

q = 25i + (k + 4)j = 25i + ( -6/17 + 4)j

q = 25i + 62/17 j

|q| = √ 25² + (62/17)²

= 25.2646

Now applying them to the formula above

We already have p.q = 11 from the question.

11 = (1.0424)(25.2646)Cos∅

11 = 26.3358Cos∅

Cos∅ = 11/26.3358

Cos∅ = 0.4177

∅ = 65.31°.

This should be it.

Hope it helps

6 0

3 years ago

5d/9 = 10 What does d equal? (Please explain)

Semenov [28]

Answer:

d=10

Step-by-step explanation:

first you divide 5d by 5 to cancel it self out and make it only d.

Then divide 10 by 5.

you should then get 2

5 0

2 years ago

Write the sentence as an equation.<br> 102 is the same as the product of 189 and p, decreased by 80

rosijanka [135]

Answer:

102+189p-80

Step-by-step explanation:

8 0

3 years ago

What are the orders of 3,7,9,11,13,17 and 19(mod20)?does 20 have primitive roots?

bezimeni [28]

$3\equiv3\mod{20}$
$3^2\equiv9\mod{20}$
$3^3\equiv27\equiv7\mod{20}$
$3^4\equiv3\cdot3^3\equiv3\cdot7\equiv21\equiv1\mod{20}$

$7\equiv7\mod{20}$
$7^2\equiv49\equiv9\mod{20}$
$7^3\equiv7\cdot7^2\equiv63\equiv3\mod{20}$
$7^4\equiv7\cdot7^3\equiv21\equiv1\mod{20}$

$9\equiv9\mod{20}$
$9^2\equiv3^4\equiv1\mod{20}$

$11\equiv11\mod{20}$
$11^2\equiv121\equiv1\mod{20}$

$13\equiv-7\equiv13\mod{20}$
$13^2\equiv169\equiv9\mod{20}$
$13^3\equiv13\cdot13^2\equiv(-7)9\equiv-63\equiv-3\mod{20}$
$13^4\equiv13\cdot13^3\equiv(-7)(-3)\equiv21\equiv1\mod{20}$

$17\equiv-3\equiv17\mod{20}$
$17^2\equiv(-3)^2\equiv9\mod{20}$
$17^3\equiv(-3)^3\equiv-27\equiv3\mod{20}$
$17^4\equiv(-3)^4\equiv81\equiv1\mod{20}$

$19\equiv-1\equiv19\mod{20}$
$19^2\equiv19(-1)\equiv-19\equiv1\mod{20}$

Generally speaking, a number $x$ coprime to $n$ will be a primitive root of $n$ if we have $x^n\equiv x\mod{n}$ , or $x^{n-1}\equiv1\mod{n}$ . In other words, if $x$ is of order $n-1$ modulo $n$ , then $x$ is a primitive root of $n$ .

Since none of these numbers has order 19, it follows that 20 does not have any primitive roots.

6 0

4 years ago

Plz give me answer anybody fast​

Plz give me answer anybody fast