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3 years ago

Draw the translation.

Mathematics

2 answers:

Brums [2.3K]3 years ago

8 0

Answer:

of what this Desenați traducerea

Step-by-step explanation:

shutvik [7]3 years ago

6 0

Answer:

G(-3,-2) W(-3,0) E(1,1)

Step-by-step explanation:

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A triangle has sides of length 11 m, 12 m, and 16 m.

Hunter-Best [27]

We know that
the law of cosines formulas established
c²=a²+b²-2*a*b*cos C-----> cos C=[a²+b²-c²]/[2*a*b]
a=11 m
b=12 m
c=16 m
so
cos C=[11²+12²-16²]/[2*11*12]---> cos C=0.0341
C=arc cos (0.0341)-------> C=88.05°

see the attached figure

the answer is
the angle is 88°

8 0

4 years ago

Read 2 more answers

Find the first 4 terms of the sequence defined by this rule: f(n)=4n-7

tatiyna

Answer:

-3, 1, 5, 9

Step-by-step explanation:

f(n) = 4n - 7

f(1) = 4(1) - 7

f(1) = 4 - 7

f(1) = -3

f(2) = 4(2) - 7

f(2) = 8 - 7

f(2) = 1

f(3) = 4(3) - 7

f(3) = 12 - 7

f(3) = 5

f(4) = 4(4) - 7

f(4) = 16 - 7

f(4) = 9

6 0

3 years ago

1.5 more than the quotient of m and 5 is g

seropon [69]

You can do it! :) I believe in you

3 0

3 years ago

Evaluate P(6,2).<br> A.) 12<br> B.) 30<br> C.) 360

Otrada [13]

Permutation
$6P2=\frac{6!}{(6-2)!}=\frac{4!\cdot 5 \cdot 6}{4!}=5 \cdot 6=30$

7 0

3 years ago

Read 2 more answers

The sum of the series {1(2/3)}²+{2(1)3)}²+3²+{3(2/3)}²+....to 10 term is

ryzh [129]

Step-by-step explanation:

<h3><u>Given Question </u></h3>

The sum of the series is

$\tt{ {\bigg[1\dfrac{2}{3} \bigg]}^{2} + {\bigg[2\dfrac{1}{3} \bigg]}^{2} + {3}^{2} + {\bigg[3\dfrac{2}{3} \bigg]}^{2} + - - 10 \: terms}$

$\green{\begin{gathered}\large{\sf{{\underline{Formula \: Used - }}}} \end{gathered}}$

$\boxed{\tt{ \displaystyle\sum_{k=1}^{n}1 = n \: }}$

$\boxed{\tt{ \displaystyle\sum_{k=1}^{n}k = \frac{n(n + 1)}{2} \: }}$

$\boxed{\tt{ \displaystyle\sum_{k=1}^{n} {k}^{2} = \frac{n(n + 1)(2n + 1)}{6} \: }}$

$\large\underline{\sf{Solution-}}$

Given series is

$\rm :\longmapsto\: {\bigg[1\dfrac{2}{3} \bigg]}^{2} + {\bigg[2\dfrac{1}{3} \bigg]}^{2} + {3}^{2} + {\bigg[3\dfrac{2}{3} \bigg]}^{2} + - - - 10 \: terms$

can be rewritten as

$\rm \: = \: {\bigg[\dfrac{5}{3} \bigg]}^{2} + {\bigg[\dfrac{7}{3} \bigg]}^{2} + {\bigg[\dfrac{9}{3} \bigg]}^{2} + {\bigg[\dfrac{11}{3} \bigg]}^{2} + - - - 10 \: terms$

$\rm \: = \: \dfrac{1}{9}[ {5}^{2} + {7}^{2} + {9}^{2} + - - - 10 \: terms \: ]$

Now, here, 5, 7, 9 forms an AP series with first term 5 and common difference 2.

So, its general term is given by 5 + ( n - 1 )2 = 5 + 2n - 2 = 2n + 3

So, above series can be represented as

$\rm \: = \: \dfrac{1}{9}\displaystyle\sum_{n=1}^{10}(2n + 3) ^{2}$

$\rm \: = \: \dfrac{1}{9}\displaystyle\sum_{n=1}^{10}\bigg[ {4n}^{2} + 9 + 12n\bigg]$

$\rm \: = \: \dfrac{1}{9}\bigg[\displaystyle\sum_{n=1}^{10} {4n}^{2} + \displaystyle\sum_{n=1}^{10}9 + 12\displaystyle\sum_{n=1}^{10}n\bigg]$

$\rm \: = \: \dfrac{1}{9}\bigg[4\displaystyle\sum_{n=1}^{10} {n}^{2} +9 \displaystyle\sum_{n=1}^{10}1 + 12\displaystyle\sum_{n=1}^{10}n\bigg]$

$\rm \: = \: \dfrac{4}{9}\bigg[\dfrac{10(10 + 1)(20 + 1)}{6} \bigg] + 10 + \dfrac{4}{3}\bigg[\dfrac{10(10 + 1)}{2} \bigg]$

$\rm \: = \: \dfrac{4}{9}\bigg[\dfrac{10(11)(21)}{6} \bigg] + 10 + \dfrac{4}{3}\bigg[\dfrac{10(11)}{2} \bigg]$

$\rm \: = \: \dfrac{1540}{9} + 10 + \dfrac{220}{3}$

$\rm \: = \: \dfrac{1540 + 90 + 660}{9}$

$\rm \: = \: \dfrac{2290}{9}$

Hence,

$\boxed{\tt{ {\bigg[1\dfrac{2}{3} \bigg]}^{2} + {\bigg[2\dfrac{1}{3} \bigg]}^{2} + {3}^{2} + {\bigg[3\dfrac{2}{3} \bigg]}^{2} + - - 10 \: terms = \frac{2290}{9}}}$

6 0

3 years ago