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3 years ago

Do you guys know how to remember certain measurements?

1 answer:

6 0

For what measurements? if you’re talking like science stuff i use King Henry Drank My Delicious Chocolate Milk for Kilo Hecto Decka, middle, Deci, Centi, Milli. middle is your unit!

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Find the value of the lettered angles in each of the following ( see image ).

Arisa [49]

Step-by-step explanation:

here's the solution :-

1.) =》f = 90°

( because angle made by diameter on circumference of the circle measures 90° )

2.) =》here,

g + f + 56° = 180°

( because the sum of all interior angles of the triangle measures 180° )

=》g + 90° + 56° = 180°

=》g + 146° = 180°

=》g = 180° - 146°

=》g = 34°

3.) h = 56°

( since, they form alternate interior pair which are equal )

4 0

3 years ago

Read 2 more answers

Cual es la suma de los ángulos externos de un triángulo

mr_godi [17]

“what is the sum of the angles of a triangle” 180 degrees

4 0

3 years ago

3. Find the difference between the two expressions. 1x - 5y)-1(x + 2y)

Fittoniya [83]

Answer:

$-7y$

Step-by-step explanation:

Our equation is:

$\left(1x-5y\right)-1\left(x+2y\right)$

First we dritribute

Now we have, $x-5y-x-2y$

$Combine$ light terms

and we get $-7y$

5 0

3 years ago

Harry rode 2 1/2 miles on his bike yesterday. He rode 1 3/4 miles on his bike today. If he plans to ride 6 miles this week on hi

aksik [14]

Answer:

1 3/4 miles

Step-by-step explanation:

2 1/2+ 1 3/4= 4 1/4

6-4 1/4= 1 3/4

3 0

3 years ago

Read 2 more answers

Find the vector projection of B onto A if A = 5i + 11j – 2k,B = 4i + 7k

valkas [14]

Answer:

$\frac{3}{13} i+\frac{11}{130} j-\frac{6}{65} k\\$

Step-by-step explanation:

Given A = 5i + 11j – 2k and B = 4i + 7k, the vector projection of B unto a is expressed as $proj_ab = \dfrac{b.a}{||a||^2} * a$

b.a = (5i + 11j – 2k)*( 4i + 0j + 7k)

note that i.i = j.j = k.k =1

b.a = 5(4)+11(0)-2(7)

b.a = 20-14

b.a = 6

||a|| = √5²+11²+(-2)²

||a|| = √25+121+4

||a|| = √130

square both sides

||a||² = (√130)

||a||² = 130

$proj_ab = \dfrac{6}{130} * (5i+11j-2k)\\\\proj_ab = \frac{30}{130} i+\frac{11}{130} j-\frac{12}{130} k\\\\proj_ab = \frac{3}{13} i+\frac{11}{130} j-\frac{6}{65} k\\\\$

Hence the projection of b unto a is expressed as $\frac{3}{13} i+\frac{11}{130} j-\frac{6}{65} k\\$

7 0

3 years ago