All categories

3 years ago

Show (with tens and hundreds) how to count from 60 to 430

Mathematics

1 answer:

Rzqust [24]3 years ago

8 0

Answer:

three groups of 100 and 7 groups of 10

Step-by-step explanation:

430-60=370

370=3x100 + 7x10

You might be interested in

Two angles are supplementary, and one of the measures 35 degrees. What is the measure of the other angle?

Alekssandra [29.7K]

Answer=145

this is because supplementary means the two angles equal 180
180-35=145

3 0

2 years ago

Edward can run 1/2 mile in 300 seconds. What is his unit of rate

sattari [20]

Im not quite sure but i think the unit rate is 150. Hope this helps!!

5 0

3 years ago

Which statements are true? Mark all that apply.

FrozenT [24]

Answer:

A and D

Step-by-step explanation:

3 0

3 years ago

Solve for the missing side. Round to the nearest tenth.<br> х<br> 51°<br> 12

alisha [4.7K]

9514 1404 393

Answer:

7.6

Step-by-step explanation:

The relevant relation is ...

Cos = Adjacent/Hypotenuse

cos(51°) = x/12

x = 12·cos(51°) ≈ 7.552

x ≈ 7.6

6 0

3 years ago

Find x?<br> In 3x - In(x - 4) = ln(2x - 1) +ln3

earnstyle [38]

Answer:

$x = \displaystyle \frac{5 + \sqrt{17}}{2}$ .

Step-by-step explanation:

Because $3\, x$ is found in the input to a logarithm function in the original equation, it must be true that $3\, x > 0$ . Therefore, $x > 0$ .

Similarly, because $(x - 4)$ and $(2\, x - 1)$ are two other inputs to the logarithm function in the original equation, they should also be positive. Therefore, $x > 4$ .

Let $a$ and $b$ represent two positive numbers (that is: $a > 0$ and $b > 0$ .) The following are two properties of logarithm:

$\displaystyle \ln (a) + \ln(b) = \ln\left(a \cdot b\right)$ .

$\displaystyle \ln (a) - \ln(b) = \ln\left(\frac{a}{b}\right)$ .

Apply these two properties to rewrite the original equation.

Left-hand side of this equation:

$\begin{aligned}&\ln(3\, x) - \ln(x - 4)= \ln\left(\frac{3\, x}{x -4}\right)\end{aligned}$

Right-hand side of this equation:

$\ln(2\, x- 1) + \ln(3) = \ln\left(3 \left(2\, x - 1\right)\right)$ .

Equate these two expressions:

$\begin{aligned}\ln\left(\frac{3\, x}{x -4}\right) = \ln(3(2\, x - 1))\end{aligned}$ .

The natural logarithm function $\ln$ is one-to-one for all positive inputs. Therefore, for the equality $\begin{aligned}\ln\left(\frac{3\, x}{x -4}\right) = \ln(3(2\, x - 1))\end{aligned}$ to hold, the two inputs to the logarithm function have to be equal and positive. That is:

$\displaystyle \frac{3\ x}{x - 4} = 3\, (2\, x - 1)$ .

Simplify and solve this equation for $x$ :

$x^2 - 5\, x + 2 = 0$ .

There are two real (but not rational) solutions to this quadratic equation: $\displaystyle \frac{5 + \sqrt{17}}{2}$ and $\displaystyle \frac{5 - \sqrt{17}}{2}$ .

However, the second solution, $\displaystyle \frac{5 - \sqrt{17}}{2}$ , is not suitable. The reason is that if $x = \displaystyle \frac{5 - \sqrt{17}}{2}$ , then $(x - 4)$ , one of the inputs to the logarithm function in the original equation, would be smaller than zero. That is not acceptable because the inputs to logarithm functions should be greater than zero.

The only solution that satisfies the requirements would be $\displaystyle \frac{5 + \sqrt{17}}{2}$ .

Therefore, $x = \displaystyle \frac{5 + \sqrt{17}}{2}$ .

7 0

3 years ago