All categories

3 years ago

Which expression is 5 times as large as the expression 345+23

Mathematics

2 answers:

anyanavicka [17]3 years ago

6 0

Answer: 1840

Step-by-step explanation:

you add 345 + 23 and multiply by 5

natima [27]3 years ago

3 0

Answer:

1840

Step-by-step explanation:

345+23=368*5

You might be interested in

5 - 4x = 17 Please help meeeeeeeee

scZoUnD [109]

Answer: x = -3

Step-by-step explanation:

5 - 4x = 17

Solve:

-4x = 17 - 5

-4x = 12

x = 12/(-4)

x = -3

Therefore, the answer is -3.

5 0

3 years ago

Read 2 more answers

<img src="https://tex.z-dn.net/?f=2%20%7C3%20%5Csqrt%7B2%7D%20-%202%20%5Csqrt%7B3%7D%20%20%7C%20%20%2B%20%20%7C3%20%20%5Csqrt%7B

kiruha [24]

Answer:

hope it helps

Step-by-step explanation:

7 0

3 years ago

On a particular day, the wind added 3 km per hour to Alfonso's rate when he was cycling with the wind and subtracted 3 km per ho

Vesnalui [34]

Answer:

31 km?

Step-by-step explanation:

3 0

3 years ago

How tall is the tower? 48° 16 ft -30 ft [?] ft Round to the nearest tenth.

Marina86 [1]

Answer:

49.3 ft

Step-by-step explanation:

Using the tangent ratio in the right triangle

tan48° = $\frac{opposite}{adjacent}$ = $\frac{x}{30}$ ( multiply both sides by 30 )

30 × tan48° = x , then

x ≈ 33.3

Height of tower = 33.3 + 16 = 49.3 ft ( to the nearest tenth )

4 0

2 years ago

Read 2 more answers

Need help with this math question

anastassius [24]

Answer:

The vertex is: $(6, 8)$

Step-by-step explanation:

First solve the equation for the variable y

$x^2-4y-12x+68=0$

Add 4y on both sides of the equation

$4y=x^2-4y+4y-12x+68$

$4y=x^2-12x+68$

Notice that now the equation has the general form of a parabola

$ax^2 +bx +c$

In this case

$a=1\\b=-12\\c=68$

Add $(\frac{b}{2}) ^ 2$ and subtract $(\frac{b}{2}) ^ 2$ on the right side of the equation

$(\frac{b}{2}) ^ 2=(\frac{-12}{2}) ^ 2\\\\(\frac{b}{2}) ^ 2=(-6) ^ 2\\\\(\frac{b}{2}) ^ 2=36$

$4y=(x^2-12x+36)-36+68$

Factor the expression that is inside the parentheses

$4y=(x-6)^2+32$

Divide both sides of the equality between 4

$\frac{4}{4}y=\frac{1}{4}(x-6)^2+\frac{32}{4}$

$y=\frac{1}{4}(x-6)^2+8$

For an equation of the form

$y=a(x-h)^2 +k$

the vertex is: (h, k)

In this case

$h=6\\k =8$

the vertex is: $(6, 8)$

5 0

3 years ago