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3 years ago

14

Louisa needs 3 liters of lemonade and punch for a picnic she has 1800 milliliters of lemonade. How much punch does she need expl

ain how you found your answer

2 answers:

Olin [163]3 years ago

7 0

If 1800 ml is 1.8 litres, do 3-1.8=1.2

8090 [49]3 years ago

5 0

There are 1000 millimetres in a litre, so 1800 ml is 1.8 litres. 3-1.8 is 1.2.

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Find the exact value of cos(a+b) if cos a=-1/3 and cos b=-1/4 if the terminal side if a lies in quadrant 3 and the terminal side

maria [59]

Answer:

cos(a + b) = $\frac{1}{12}(1-2\sqrt{30})$

Step-by-step explanation:

cos(a + b) = cos(a).cos(b) - sin(a).sin(b) [Identity]

cos(a) = $-\frac{1}{3}$

cos(b) = $-\frac{1}{4}$

Since, terminal side of angle 'a' lies in quadrant 3, sine of angle 'a' will be negative.

sin(a) = $-\sqrt{1-(-\frac{1}{3})^2}$ [Since, sin(a) = $\sqrt{(1-\text{cos}^2a)}$ ]

= $-\sqrt{\frac{8}{9}}$

= $-\frac{2\sqrt{2}}{3}$

Similarly, terminal side of angle 'b' lies in quadrant 2, sine of angle 'b' will be negative.

sin(b) = $-\sqrt{1-(-\frac{1}{4})^2}$

= $-\sqrt{\frac{15}{16}}$

= $-\frac{\sqrt{15}}{4}$

By substituting these values in the identity,

cos(a + b) = $(-\frac{1}{3})(-\frac{1}{4})-(-\frac{2\sqrt{2}}{3})(-\frac{\sqrt{15}}{4})$

= $\frac{1}{12}-\frac{\sqrt{120}}{12}$

= $\frac{1}{12}(1-\sqrt{120})$

= $\frac{1}{12}(1-2\sqrt{30})$

Therefore, cos(a + b) = $\frac{1}{12}(1-2\sqrt{30})$

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3 years ago

In quadrilateral QRST, . Is QRST a rectangle?

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Is there a picture with it?

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3 years ago

I need help finding the area

aniked [119]

<h3>Given :</h3>

Base of triangle = 7 yd
Height of triangle = 10 yd

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<h3>To find:</h3>

Area of triangle

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We know:-

When base and height of triangle is given we use this formula:

$\bigstar \boxed{ \rm Area \: of \: triangle = \frac{base \times height}{2} }$

$\\ \\$

So:-

$\\ \\$

$\dashrightarrow \sf \: Area \: of \: triangle = \dfrac{base \times height}{2} \\$

$\\ \\$

$\dashrightarrow \sf \: Area \: of \: triangle = \dfrac{7 \times 10}{2} \\$

$\\ \\$

$\dashrightarrow \sf \: Area \: of \: triangle = \dfrac{7 \times 5 \times2 }{2} \\$

$\\ \\$

$\dashrightarrow \sf \: Area \: of \: triangle = \dfrac{7 \times 5 \times\cancel2 }{\cancel2} \\$

$\\ \\$

$\dashrightarrow \sf \: Area \: of \: triangle = \dfrac{7 \times 5 \times1 }{1} \\$

$\\ \\$

$\dashrightarrow \sf \: Area \: of \: triangle =7 \times 5$

$\\ \\$

$\dashrightarrow \bf \: Area \: of \: triangle =35 {yd}^{2} \\$

$\\ \\$

$\therefore \underline{\textsf{ \textbf {\: Area \: of \: triangle = \red{35}}} { \red{\bf{yd} }^{ \red2} }}$

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<h3>know more :-</h3>

$\small\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\begin {array}{cc}\\ \dag\quad \Large\underline{\bf \small{Formulas\:of\:Areas:-}}\\ \\ \star\sf Square=(side)^2\\ \\ \star\sf Rectangle=Length\times Breadth \\\\ \star\sf Triangle=\dfrac{1}{2}\times Base\times Height \\\\ \star \sf Scalene\triangle=\sqrt {s (s-a)(s-b)(s-c)}\\ \\ \star \sf Rhombus =\dfrac {1}{2}\times d_1\times d_2 \\\\ \star\sf Rhombus =\:\dfrac {1}{2}d\sqrt {4a^2-d^2}\\ \\ \star\sf Parallelogram =Base\times Height\\\\ \star\sf Trapezium =\dfrac {1}{2}(a+b)\times Height \\ \\ \star\sf Equilateral\:Triangle=\dfrac {\sqrt{3}}{4}(side)^2\end {array}}\end{gathered}\end{gathered}\end{gathered}$ ]

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2 years ago