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4 years ago

Please help me 4 1/7 + 1/2

Mathematics

2 answers:

zmey [24]4 years ago

8 0

Answer:

Exact Form:65/14

Mixed number Form:4 9/14

Step-by-step explanation:

Alex73 [517]4 years ago

4 0

Answer:

4 9/14

as an improper fraction: 65/14

hopefully this helped :3

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Find the area of the trapezoid below.

Pani-rosa [81]

Answer: 54 cm²

Step-by-step explanation: In this problem, we're asked to find the area of the trapezoid shown. A trapezoid is a quadrilateral with one pair of parallel sides.

The formula for the area of a trapezoid is shown below.

$Area =\frac{1}{2} (^{b} 1 + ^{b}2)h$

The b's represent the bases which are the parallel sides and h is the height.

So in the trapezoid shown, the bases are 6 cm and 12 cm and the height is 6 cm. Plugging this information into the formula, we have $\frac{1}{2} (6 cm +12cm)(6 cm)$ .

Next, the order of operations tell us that we must simplify inside the parentheses first. 6 cm + 12 cm is 18 cm and we have $\frac{1}{2}(18 cm)(6 cm)$ .

$\frac{1}{2} (18 cm)$ is 9 cm and we have 9 cm · 6 cm of 54 cm²

So the area of the trapezoid shown is 54 cm².

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3 years ago

Which ordered pair is not a solution to the equation y = 2x?

Yuliya22 [10]

Answer:

Step-by-step explanation:

Y=1/2y

8 0

3 years ago

Find the magnitude of the net electric field at the origin (created by charges q1 and q2).

Readme [11.4K]

The direction of the net magnetic field is $167.36^{\circ}$

<h3>How to find the magnitude?</h3>

$q_1=-4 \mathrm{nC} \text { at }(0.6,0.8)$

$q_2=6 \mathrm{nC}$ at (0.6,0)

$r_1=$ Distance of $q_1$ from origin $=\sqrt{0.6^2+0.8^2}$

$r_2=$ Distance of $q_2$ from origin $=0.6$

$\mathrm{k}=$ Coulomb constant $$=9$ times $10^9 \mathrm{Nm}^2 / \mathrm{C}^2$$

Electric field is given by

$&E_1=\frac{k q_1}{r_1^2} \\$

$&\Rightarrow E_1=\frac{9 \times 10^9 \times 4 \times 10^{-9}}{\sqrt{0.6^2+0.8^2}} \\$

$&\Rightarrow E_1=36 \mathrm{~N} / \mathrm{C}$

$&\theta=\tan ^{-1} \frac{0.8}{0.6} \\$

$&\Rightarrow \theta=53.13^{\circ} \\$

$&E_1=36 \cos 53.13^{\circ} \hat{i}+36 \sin 53.13^{\circ} \hat{j} \\$

$&\Rightarrow E_1=21.6 \hat{i}+28.8 \hat{j}$

$&E_2=\frac{k q_2}{r_2^2} \\$

$&\Rightarrow E_2=\frac{9 \times 10^9 \times 6 \times 10^{-9}}{0.6^2} \\$

$&\Rightarrow E_2=150 \mathrm{~N} / \mathrm{C}$

The charge $q_2$ is on the $\mathrm{x}$ axis itself and it is pointing towards the origin (left side) so the sign will be negative

$E_2=-150 \hat{i}$

Resultant electric field

$&E=E_1+E_2 \\$

$&\Rightarrow E=21.6 \hat{i}+28.8 \hat{j}+(-150 \hat{i}) \\$

$&\Rightarrow E=-128.4 \hat{i}+28.8 \hat{j}$

Magnitude of electric field is given by

$&|E|=\sqrt{(-128.4)^2+28.8^2} \\$

$&\Rightarrow|E|=121.6 \mathrm{~N} / \mathrm{C}$

Magnitude of the net electric field is $121.6 \mathrm{~N} / \mathrm{C}$

Direction is given by

$\theta=\tan ^{-1} \frac{128.4}{28.8}=77.36^{\circ}$

From the -x axis

$(90+77.36)^{\circ}=167.36^{\circ}$

The direction of the net magnetic field is $167.36^{\circ}$

To learn more about magnetic field, refer to:

brainly.com/question/26257705

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6 0

2 years ago

When you graduate college at the age of 20, you want to start saving up for retirement. If your investment pays a fixed APR of 9

Dominik [7]

Answer-

You need to deposit $337.62 each month, to reach this goal.

Solution-

We know that,

$\text{FV of annuity}=P[\frac{(1+r)^n-1}{r}]$

Where,

P = periodic payment

r = rate per period

n = number of period

Here,

$FV\ of\ annuity=2,500,000,\\\\P=?,\\\\r = 9\%\ annually=\frac{9}{12}\%\ monthly=\frac{9}{1200}\ monthly\\\\n=45\ years=45\times 12=540\ months$

Putting the values,

$\Rightarrow 2500000=P[\frac{(1+\frac{9}{1200})^{540}-1}{{\frac{9}{1200}}}]$

$\Rightarrow P=\frac{2500000}{[\frac{(1+\frac{9}{1200})^{540}-1}{{\frac{9}{1200}}}]}$

$\Rightarrow P=\frac{2500000}{\frac{56.5365-1}{0.0075}}$

$\Rightarrow P=\frac{2500000}{\frac{55.5365}{0.0075}}$

$\Rightarrow P=337.62$

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3 years ago

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Which point on the x-axis lies on the line that passes through point C and is parallel to line AB?

sweet [91]

It is ( 2,0 )
It took me a while because I didn't went to school for 2 months.

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3 years ago

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