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3 years ago

A) Write 7.1 x 10² as an ordinary number.

Mathematics

2 answers:

Mkey [24]3 years ago

6 0

710

Move the decimal two spaces to the right

Vilka [71]3 years ago

4 0

Answer:

710 is the answer.

Step-by-step explanation:

7.1 x 10^2

10^2= 10 x 10 = 100

so,

7.1 x 100 = 710

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Which function is the inverse of 6) = 2x+39<br> ASAP

sergij07 [2.7K]

Answer:

The inverse is 1/2x - 3/2

Step-by-step explanation:

y = 2x+3

To find the inverse, exchange x and y

x = 2y+3

Solve for y

x-3 =2y+3-3

x-3 = 2y

Divide by 2

1/2x - 3/2 = 2y/2

1/2x - 3/2 = y

The inverse is 1/2x - 3/2

7 0

2 years ago

Which expressions correctly represent the area of the triangle XYZ?

Evgen [1.6K]

Answer:

Options (2) and Option (1)

Step-by-step explanation:

Area of the ΔXYZ = $\frac{1}{2}(\text{Base})(\text{Height})$

= $\frac{1}{2}(XZ)h$

= $\frac{1}{2}(18)h$

By applying sine rule in the ΔYPZ,

sin(38°) = $\frac{\text{Opposite side}}{\text{Hypotenuse}}$

sin(38°) = $\frac{h}{14}$

h = 14[sin(38°)]

Therefore, area of the given triangle = $\frac{1}{2}(18)(14)sin(38)$

By applying sine rule in ΔYPZ,

sin(58°) = $\frac{\text{Opposite side}}{\text{Hypotenuse}}$

sin(58°) = $\frac{h}{10}$

h = 10sin(58°)

Area of ΔXYZ = $\frac{1}{2}(18)(10)sin(58)$

Therefore, Options (2) and Option (1) are the correct options.

6 0

3 years ago

Each investment matures in 3 years. The interest compounds annually.

Sveta_85 [38]

bearing in mind that 4¾ is simply 4.75.

$\bf ~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\$600\\ r=rate\to 5\%\to \frac{5}{100}\dotfill &0.05\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{annually, thus once} \end{array}\dotfill &1\\ t=years\dotfill &3 \end{cases} \\\\\\ A=600\left(1+\frac{0.05}{1}\right)^{1\cdot 3}\implies A=600(1.05)^3\implies A=694.575 \\\\[-0.35em] ~\dotfill$

$\bf ~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\$750\\ r=rate\to 4.75\%\to \frac{4.75}{100}\dotfill &0.0475\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{annually, thus once} \end{array}\dotfill &1\\ t=years\dotfill &3 \end{cases} \\\\\\ A=750\left(1+\frac{0.0475}{1}\right)^{1\cdot 3}\implies A=750(1.0475)^3\implies A\approx 862.032$