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2 years ago

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Pls giys just anser this question I just need help and this app is for helping so please write it no files ok because I know it’

s a bot

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Naya [18.7K]2 years ago

6 0

Answer:

please can I get the full picture I can't fully understand

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Molly tried to evaluate 93\times5193×5193, times, 51 using partial products. Her work is shown below.

Mrac [35]

Since molly's solution tally's with the given solution, hence <em>Molly's solution is correct.</em>

Given the working on a partial product of 93 and 51 carried out by Molly as shown:

$\begin{array}{llrr} &&93 \\ &&\underline{{}\times51} \\ &\blueD{\text{Step 1}}&\blueD{3}& \blueD{1\times3\text{ ones}}\\ &\greenD{\text{Step 2}}&\greenD{90}& \greenD{1\times 9\text{ tens}}\\ &\maroonD{\text{Step 3}}&\maroonD{150}& \maroonD{50\times 3\text{ ones}}\\ &\goldE{\text{Step 4}}&\underline{{}+\goldE{ 4{,}500}}& \goldE{50\times 9\text{ tens}}\\ &\purpleD{\text{Step 5}}&\purpleD{4{,}743}& \end{array}$

This partial product can also be solved as shown below:

$93 \times 51 = (90+3)\times (50+1)$

Applying the distributive law:

$93 \times 51 = 90(50) + 90(1) + 3(50) + 3(1)\\93 \times 51 =4500 + 90 + 150 + 3\\93 \times 51 =4500+240+3\\93 \times 51 =4740+3\\93 \times 51 =4743$

Since molly's solution tally's with the given solution, hence <em>Molly solution is correct.</em>

Learn more about partial product at: brainly.com/question/24716925

5 0

2 years ago

I HAVE 2 MIN PLS HELP

Maksim231197 [3]

Answer:

C (ignore this it asked for 20 characters)

5 0

3 years ago

Express the area A of a circle as a function of its radius, r. Determine the area of a circle whose radius is 11 inches. That

nekit [7.7K]

<h2>Answer with explanation:</h2>

The function represents the area of a circle(A) is given by :-

$A(r)=2\pi r$

To determine the area of a circle whose radius is 11 inches. , we put the value of r=11 inches, we get

$A(11)=2\pi (11)$

Now, we put $\pi=\dfrac{22}{7}$ in the above function , we get

$A(11)=2(\dfrac{22}{7} )(11)\\\\\Rightarrow\ A=56.5714285714\approx56.57$

8 0

3 years ago

How do I solve this problem?

Ainat [17]

O to f will be 54-59

6 0

3 years ago

Read 2 more answers

Given: ABC is a right triangle with right angle C. AC=15 centimeters and m∠A=40∘ . What is BC ? Enter your answer, rounded to th

konstantin123 [22]

In order to answer this question, the figure in the first picture will be helpful to understand what a right triangle is. Here, a right angle refers to $90\°$ .

However, if we want to solve the problem we have to know certain things before:

In the second figure is shown a general right triangle with its three sides and another given angle, we will name it $\alpha$ :

The side <u>opposite to the right angle</u> is called The Hypotenuse (h)

The side <u>opposite to the angle $\alpha$ </u> is called the Opposite (O)

The side <u>next to the angle $\alpha$ </u> is called the Adjacent (A)

So, going back to the triangle of our question (first figure):

The Hypotenuse is AB

The Opposite is BC

The Adjacent is AC

Now, if we want to find the length of each side of a right triangle, we have to use the <u>Pythagorean Theorem</u> and T<u>rigonometric Functions:</u>

Pythagorean Theorem

$h^{2}=A^{2} +O^{2}$ (1)

Trigonometric Functions (here are shown three of them):

Sine: $sin(\alpha)=\frac{O}{h}$ (2)

Cosine: $cos(\alpha)=\frac{A}{h}$ (3)

Tangent: $tan(\alpha)=\frac{O}{A}$ (4)

In this case the function that works for this problem is cosine (3), let’s apply it here:

$cos(40\°)=\frac{AC}{h}$

$cos(40\°)=\frac{15}{h}$ (5)

And we will use the Pythagorean Theorem to find the hypotenuse, as well:

$h^{2}=AC^{2}+BC^{2}$

$h^{2}=15^{2}+BC^{2}$ (6)

$h=\sqrt{225+BC^2}$ (7)

Substitute (7) in (5):

$cos(40\°)=\frac{15}{\sqrt{225+BC^2}}$

Then clear BC, which is the side we want:

${\sqrt{225+BC^2}}=\frac{15}{cos(40\°)}$

${{\sqrt{225+BC^2}}^2={(\frac{15}{cos(40\°)})}^2$

$225+BC^{2}=\frac{225}{{(cos(40\°))}^2}$

$BC^2=\frac{225}{{(cos(40\°))}^2}-225$

$BC=\sqrt{158,41}$

$BC=12.58$

Finally $BC$ is approximately 13 cm

7 0

3 years ago

Read 2 more answers