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3 years ago

Which of the filling units are units of length

Mathematics

1 answer:

dimaraw [331]3 years ago

3 0

Answer:

The SI unit for length is the meter (abbreviated m); its definition has also changed over time to become more accurate and precise. The meter was first defined in 1791 as 1/10,000,000 of the distance from the equator to the North Pole.

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What is the point-slope form of a line that has a slope of –4 and passes through point (–3, 1)?

ivolga24 [154]

Answer:

y-1= -4(x-(-3))

y minus 1 = negative 4 left-bracket x minus (negative 3) right-bracket

Step-by-step explanation:

The formula for point slope form is written: $y-y_{1} = m(x-x_{1})$ . Now we fill in our know information. m=-4 (because that is the slope), $y_{1}$ =1 (It is the y value in the ordered pair), $x_{1}$ = -3 (It is the x value in the ordered pair).

y-1=-4(x- (-3))

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3 years ago

Given $a \equiv 1 \pmod{7}$, $b \equiv 2 \pmod{7}$, and $c \equiv 6 \pmod{7}$, what is the remainder when $a^{81} b^{91} c^{27}$

Blizzard [7]

$\begin{cases}a\equiv1\pmod7\\b\equiv2\pmod7\\c\equiv6\pmod7\end{cases}$

$a^{81}\equiv1^{81}\equiv1\pmod7$

$b^{91}\equiv2^{91}\pmod7$

$2^{91}\equiv(2^3)^{30}\times2^1\equiv8^{30}\times2\pmod7$
$8\equiv1\pmod7$
$2\equiv2\pmod7$
$\implies2^{91}\equiv1^{30}\times2\equiv2\pmod7$

$c^{27}\equiv6^{27}\pmod7$

$6^{27}\equiv(-1)^{27}\equiv-1\equiv6\pmod7$

$\implies a^{81}b^{91}c^{27}\equiv1\times2\times6\equiv12\equiv5\pmod7$

6 0

3 years ago

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3 years ago

Read 2 more answers

Field book of an land is given in the figure. It is divided into 4 plots . Plot I is a right triangle , plot II is an equilatera

Kazeer [188]

Answer:

Total area = 237.09 cm²

Step-by-step explanation:

Given question is incomplete; here is the complete question.

Field book of an agricultural land is given in the figure. It is divided into 4 plots. Plot I is a right triangle, plot II is an equilateral triangle, plot III is a rectangle and plot IV is a trapezium, Find the area of each plot and the total area of the field. ( use √3 =1.73)

From the figure attached,

Area of the right triangle I = $\frac{1}{2}(\text{Base})\times (\text{Height})$

Area of ΔADC = $\frac{1}{2}(\text{CD})(\text{AD})$

= $\frac{1}{2}(\sqrt{(AC)^2-(AD)^2})(\text{AD})$

= $\frac{1}{2}(\sqrt{(13)^2-(19-7)^2} )(19-7)$

= $\frac{1}{2}(\sqrt{169-144})(12)$

= $\frac{1}{2}(5)(12)$

= 30 cm²

Area of equilateral triangle II = $\frac{\sqrt{3} }{4}(\text{Side})^2$

Area of equilateral triangle II = $\frac{\sqrt{3}}{4}(13)^2$

= $\frac{(1.73)(169)}{4}$

= 73.0925

≈ 73.09 cm²

Area of rectangle III = Length × width

= CF × CD

= 7 × 5

= 35 cm²

Area of trapezium EFGH = $\frac{1}{2}(\text{EF}+\text{GH})(\text{FJ})$

Since, GH = GJ + JK + KH

17 = $\sqrt{9^{2}-x^{2}}+5+\sqrt{(15)^2-x^{2}}$

12 = $\sqrt{81-x^2}+\sqrt{225-x^2}$

144 = (81 - x²) + (225 - x²) + 2 $\sqrt{(81-x^2)(225-x^2)}$

144 - 306 = -2x² + $2\sqrt{(81-x^2)(225-x^2)}$

-81 = -x² + $\sqrt{(81-x^2)(225-x^2)}$

(x² - 81)² = (81 - x²)(225 - x²)

x⁴ + 6561 - 162x² = 18225 - 306x² + x⁴

144x² - 11664 = 0

x² = 81

x = 9 cm

Now area of plot IV = $\frac{1}{2}(5+17)(9)$

= 99 cm²

Total Area of the land = 30 + 73.09 + 35 + 99

= 237.09 cm²

7 0

3 years ago

Find the oth term of the geometric sequence 9, -18, 36, ...

Sphinxa [80]

Answer:

2304

Step-by-step explanation:

Given :-

A geometric sequence is given to us which is 9 , -18 , 36.

And we need to find out the 9th term of the sequence. Here firstly we should find the Common Ratio and then we can substitute the respective values in the formula to find the nth term of a geometric sequence .

Common Ratio :-

$:\implies$ CR = -18÷ 9 = -2

The 9 th term :-

$:\implies$ T_n = arⁿ - ¹

$:\implies$ T_9 = 9× (-2) ⁹ - ¹

$:\implies$ T_9 = 9 × (-2)⁸

$:\implies$ T_9 = 9 × 256

$:\implies$ T_9 = 2304

Hence the 10th term is 2304.

3 0

3 years ago