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2 years ago

Pls help it due ASAP Show workings Pleaseeeeeeeeeeeee

Mathematics

1 answer:

Talja [164]2 years ago

3 0

D
5 + 29 = 34/2 = 17
7 - 15 = -8/2 = -4
(17,-4)

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Write the algebraic expression that matches each graph

ArbitrLikvidat [17]

Answer:

$y=\left|x+1\right|$
$y=-\left|x\right|+1$
$y=|x|$
$y=-\left|x+1\right|+1$
$y=\left|x-2\right|-2$

Step-by-step explanation:

The graph of y=|x-k|+h is the graph of y=|x| shifted k units to the right and h units up. To reflect the equation across the x-axis (flip it upside down), the equation changes to $y=-|x|$ .

Hope this helps!

7 0

2 years ago

The nine digit number that identifies the bank that a check came from is called

patriot [66]

Answer:

The nine digit number that identifies the bank that a check came from is called The Routing Number.

5 0

3 years ago

Read 2 more answers

17cm per minute = meters per hour

Arlecino [84]

Answer: 10.2 meters per hour; or, write as: 10 ⅕ meters per hour.
___________________________________________
Explanation:
_____________________________________
Note: "h" = time in "hour(s)" ;
"cm" = length in "centimeter(s)" ;
"min" = time in "minute(s)";
"m" = length in "meter(s)"
_____________________________________
Note these EXACT conversions:
_________________________________
100 cm = 1 m ;
60 min = 1 h ;
_________________________________
To solve:

Given: 17 cm / min ; convert to: ________ m / h
__________________________________________________
(17 cm / min)* (1 m /100cm) * (60 min / 1 h) = _______ m / h
__________________________________________________
The "cm" units cancel to "1"; the "min" units cancel to "1" and we are left with units of "m/h" {"meters per hour"}.
___________________________________
We are left with (17 *1 m * 60) / (100 * 1 h) = [(17 * 60) m] / [100 h]
= [(17 *60) / 100 ] meters per hour.
________________________________________________________
To simply: (17 * 60) / 100 ;
_________________________
Method 1): To simplify: (17 * 60) / 100 ;

→Rewrite as: (17*60) / 100 = (17*20*3)/(20*5) ;
→Cancel out the "20's "; and rewrite as:

→ (17*60) / 100 = (17*3)/5 = 51/5
= 10.2 meters per hour; or, write as: 10 ⅕ meters per hour.
________________________________________________________
Method 2) To simplify: (17 * 60) / 100 ; Use calculator (or by hand):
___________________________________
→ (17 * 60) / 100 = 1,020 /100
= 10.2 meters per hour; or, write as: 10 ⅕ meters per hour.
__________________________________
Method 3) To simplify: (17 * 60) / 100 ;
_____________________________________
→ (17 * 60) / 100 = ? ;

→ Divide BOTH the "100" AND the "60" by "10";

→ 60÷10 = 6 ; 100÷10 =10; and rewrite—replace the "60" with a "6"; and replace the "100" with a "10" ;
____________________________
→ (17*60)/100 = (17*6)/10 = 102/10

= 10.2 meters per hours; or, write as: 10 ⅕ meters per hour.
____________________________________________________

7 0

3 years ago

Convert 1 cal/(m^2 * sec * °C) into BTU/(ft^2 * hr * °F)

Crazy boy [7]

Answer:

$1\ \frac{\text{cal}}{m^2\times sec\times ^\circ C}=0.03926\frac{\text{BTU}}{ft^2\times hr\times ^\circ F}$

Step-by-step explanation:

To find : Convert $1\ \frac{\text{cal}}{m^2\times sec\times ^\circ C}$ into $\frac{\text{BTU}}{ft^2\times hr\times ^\circ F}$

Solution :

We convert units one by one,

$1\text{ m}^2=10.7639\text{ ft}^2$

$1\text{ sec}=\frac{1}{3600}\text{ hour}$

$1\text{ cal}=0.003968\text{ BTU}$

Converting temperature unit,

$^\circ C\times \frac{9}{5}+32=^\circ F$

$1^\circ C\times \frac{9}{5}+32=33.8^\circ F$

So, $1^\circ C=33.8^\circ F$

Substitute all the values in the unit conversion,

$1\ \frac{\text{cal}}{m^2\times sec\times ^\circ C}=\frac{0.003968}{10.7639\times \frac{1}{3600}\times 33.8}\frac{\text{BTU}}{ft^2\times hr\times ^\circ F}$

$1\ \frac{\text{cal}}{m^2\times sec\times ^\circ C}=\frac{0.003968}{0.101061}\frac{\text{BTU}}{ft^2\times hr\times ^\circ F}$

$1\ \frac{\text{cal}}{m^2\times sec\times ^\circ C}=0.03926\frac{\text{BTU}}{ft^2\times hr\times ^\circ F}$

Therefore, The conversion of unit is $1\ \frac{\text{cal}}{m^2\times sec\times ^\circ C}=0.03926\frac{\text{BTU}}{ft^2\times hr\times ^\circ F}$

3 0

3 years ago

Use Stokes' Theorem to evaluate C F · dr F(x, y, z) = xyi + yzj + zxk, C is the boundary of the part of the paraboloid z = 1 − x

Serggg [28]

I assume $C$ has counterclockwise orientation when viewed from above.

By Stokes' theorem,

$\displaystyle\int_C\vec F\cdot\mathrm d\vec r=\iint_S(\nabla\times\vec F)\cdot\mathrm d\vec S$

so we first compute the curl:

$\vec F(x,y,z)=xy\,\vec\imath+yz\,\vec\jmath+xz\,\vec k$

$\implies\nabla\times\vec F(x,y,z)=-y\,\vec\imath-z\,\vec\jmath-x\,\vec k$

Then parameterize $S$ by

$\vec r(u,v)=\cos u\sin v\,\vec\imath+\sin u\sin v\,\vec\jmath+\cos^2v\,\vec k$

where the $z$ -component is obtained from

$1-(\cos u\sin v)^2-(\sin u\sin v)^2=1-\sin^2v=\cos^2v$

with $0\le u\le\dfrac\pi2$ and $0\le v\le\dfrac\pi2$ .

Take the normal vector to $S$ to be

$\vec r_v\times\vec r_u=2\cos u\cos v\sin^2v\,\vec\imath+\sin u\sin v\sin(2v)\,\vec\jmath+\cos v\sin v\,\vec k$

Then the line integral is equal in value to the surface integral,

$\displaystyle\iint_S(\nabla\times\vec F)\cdot\mathrm d\vec S$

$=\displaystyle\int_0^{\pi/2}\int_0^{\pi/2}(-\sin u\sin v\,\vec\imath-\cos^2v\,\vec\jmath-\cos u\sin v\,\vec k)\cdot(\vec r_v\times\vec r_u)\,\mathrm du\,\mathrm dv$

$=\displaystyle-\int_0^{\pi/2}\int_0^{\pi/2}\cos v\sin^2v(\cos u+2\cos^2v\sin u+\sin(2u)\sin v)\,\mathrm du\,\mathrm dv=\boxed{-\frac{17}{20}}$

6 0

3 years ago