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2 years ago

Estimate the weight of apples, please!

Mathematics

2 answers:

Rasek [7]2 years ago

8 0

We will solve this question by unitary method. First we will find the weight of 1 apple and then multiply the weight of 1 apple by 6 to find weight of 6 apples. Let's take the weight of 5 apples to find the weight of one apple.

$\sf \: weight \: of \: 1 \: apple = 1200 \div 5 \\ \sf \: = 240 \: grams$

<h3>Now, Weight of 6 apples</h3>

$\tt \: weight \: of \: 6 \: apples = 240 \times 6 \\ \tt = 1440 \: grams$

<em>Thus, The weight of 6 apples is 1440 grams...~</em>

Tom [10]2 years ago

4 0

Answer:

1440

Step-by-step explanation:

Hope it helps!

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I have to use trigonometric identities to solve. But I’m having trouble finding the values of cos A and sin B. Can anyone help m

katrin [286]

let's notice something, angles α and β are both in the I Quadrant, and on the first quadrant the x-coordinate/cosine and y-coordinate/sine are both positive.

$\bf \textit{Sum and Difference Identities} \\\\ cos(\alpha - \beta)= cos(\alpha)cos(\beta) + sin(\alpha)sin(\beta) \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ sin(\alpha)=\cfrac{\stackrel{opposite}{15}}{\stackrel{hypotenuse}{17}}\impliedby \textit{let's find the \underline{adjacent side}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases}$

$\bf \pm\sqrt{17^2-15^2}=a\implies \pm\sqrt{64}=a\implies \pm 8 = a\implies \stackrel{I~Quadrant}{\boxed{+8=a}} \\\\[-0.35em] ~\dotfill\\\\ cos(\beta)=\cfrac{\stackrel{adjacent}{3}}{\stackrel{hypotenuse}{5}}\impliedby \textit{let's find the \underline{opposite side}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \pm\sqrt{c^2-a^2}=b \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases}$

$\bf \pm\sqrt{5^2-3^2}=b\implies \pm\sqrt{16}=b\implies \pm 4=b\implies \stackrel{\textit{I~Quadrant}}{\boxed{+4=b}} \\\\[-0.35em] ~\dotfill$

$\bf cos(\alpha - \beta)=\stackrel{cos(\alpha)}{\left( \cfrac{8}{17} \right)}\stackrel{cos(\beta)}{\left( \cfrac{3}{5} \right)}+\stackrel{sin(\alpha)}{\left( \cfrac{15}{17} \right)}\stackrel{sin(\beta)}{\left( \cfrac{4}{5} \right)}\implies cos(\alpha - \beta)=\cfrac{24}{85}+\cfrac{60}{85} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill cos(\alpha - \beta)=\cfrac{84}{85}~\hfill$

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3 years ago

The ____ ____ separates the places larger than 1 from those that are fractions of 1, such as tenths, hundredths, etc.

polet [3.4K]

The decimal point separates the places larger than 1 from those that are fractions of 1

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3 years ago

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Whose y-intercept is (0,6) and slope is 3?

GrogVix [38]

your answer is y=3x +6

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3 years ago

If the doctor’s office is 7 units from the zoo and 2 units from the park, which ordered pair could represent the location of the

s2008m [1.1K]

Answer:

(-4, -2) is the answer

Step-by-step explanation:

is you move 7 units down from the zoo you will see ur also 2 units from the park and the coordinates would be (-4, -2)

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3 years ago

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Simplify 6m2n3 – 3m3n2 + 8m2n3 + 12m3n2

NISA [10]

Hello! Hope I can assist you.

Let us simplify some terms first!

$6m\cdot \:2n\cdot \:3 \ \textgreater \ 36mn$
$3m\cdot \:3n\cdot \:2 \ \textgreater \ 2\cdot \:3^2mn$
$8m\cdot \:2n\cdot \:3 \ \textgreater \ 48mn$
$12m\cdot \:3n\cdot \:2 \ \textgreater \ 72mn$

$36mn-2\cdot \:3^2mn+48mn+72mn$

From here, group like terms
$36mn-2\cdot \:3^2mn+48mn+72mn$

$\mathrm{Add\:similar\:elements:}\:36mn+48mn+72mn=156mn$
$156mn-2\cdot \:3^2mn$

$2\cdot \:3^2mn \ \textgreater \ 2\cdot \:9mn \ \textgreater \ 18mn$

$156mn-18mn$

$\mathrm{Add\:similar\:elements:}\:156mn-18mn=138mn$

Hope this helps!

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3 years ago