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2 years ago

9

Find the value of each expression. 42+6÷2

1 answer:

Scrat [10]2 years ago

3 0

Answer:

45

Step-by-step explanation:

6/2 = 3

42+3 = 45

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GEOMETRY: PLEASE HELP!!

Aloiza [94]

Answer:

$GF=72$

Step-by-step explanation:

All triangles in the given figure are similar, from SAS. Notice that marked in the diagram, $CD=DE=EF=FA$ .

For triangle $\triangle CGF$ was base $GF$ , leg $CF$ contains three of these marked segments. In triangle $\triangle CHE$ with base $HE$ , leg $CE$ has two of these marked segments. By definition, similar polygons have corresponding sides in a constant proportion. Therefore, the length of GF must be $3/2$ the length of EH. Since the length of EH is given as 48, we have:

$GF=\frac{3}{2}EH, \\\\GF=\frac{3}{2}\cdot 48,\\\\GF=\boxed{72}$

7 0

2 years ago

Write 125 as a product of primes. Use index notation when giving your answer.

Mnenie [13.5K]

Answer:

5^3

Step-by-step explanation:

125= 5*25

25=5*5

5 is a prime number and there are 3 of them so the answer in index notation will be 5^3

3 0

2 years ago

The area of this square

ANEK [815]

It appears as if 12cm is half the length of the side so we can assume that the total length of one side is 24cm.

The area of a square is given by the equation $A=s^{2}$ where s is the length of the side.

$A=24^{2}$

A=576

The area of this square is 576cm2.

Hope this helps!

7 0

2 years ago

If x square plus 8x plus q =(x plus k) Square find the value of k and q

PIT_PIT [208]

Answer:

The value of k is 4 and q is 16.

Step-by-step explanation:

First, you have to expand the brackets by applying :

${(a + b)}^{2} = {a}^{2} + 2ab + {b}^{2}$

${x}^{2} + 8x + q = {(x + k)}^{2}$

${x}^{2} + 8x + q = {x}^{2} + 2kx + {k}^{2}$

Next, you have to do it by comparison :

${x}^{2} + 8x + q = {x}^{2} + 2kx + {k}^{2}$

$by \: comparison$

$8x = 2kx$

$8 = 2k$

$k = 4$

$q = {k}^{2}$

$q = {4}^{2}$

$q = 16$

7 0

3 years ago

Two right triangles are shown. First, find the missing leg of the triangle on the right. Then you have two lengths to use to fin

jasenka [17]

Answer:

x = 30m

Step-by-step explanation:

<u>GIVEN :-</u>

There are 2 right-angled triangles.

In the triangle on the right side ,

Hypotenuse = 26m
Base = 10m

In the triangle on the left side ,

Base = 18m

<u>TO FIND :-</u>

Length of the common perpendicular to both triangles.
Length of the hypotenuse of the triangle on the left side.

<u>FACTS TO KNOW BEFORE SOLVING :-</u>

$Hypotenuse = \sqrt{Base^2 + Perpendicular^2}$ .
$Perpendicular = \sqrt{Hypotenuse^2 - Base^2}$

<u>SOLUTION :-</u>

In the triangle on the right side ,

$Perpendicular = \sqrt{26^2 - 10^2} = \sqrt{676 - 100} = \sqrt{576} = 24m$

⇒ Perpendicular of the triangle on right side = Perpendicular of the triangle on left side.

In the triangle on the left side ,

$Hypotenuse = \sqrt{24^2 + 18^2} = \sqrt{576 + 324} = \sqrt{900} = 30m$

∴ x = 30m

3 0

3 years ago