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2 years ago

(4h+5) and (h – 6). find the product

Mathematics

2 answers:

Andrew [12]2 years ago

7 0

Answer:

4h²– 19h – 30

Step-by-step explanation:

(4h+5)(h-6)

=4h²–24h+5h–30

=4h²–19h–30

Zepler [3.9K]2 years ago

7 0

$\huge \tt \color{pink}{A}\color{blue}{n}\color{red}{s}\color{green}{w}\color{grey}{e}\color{purple}{r }$

$\large\underline{ \boxed{ \sf{✰\:how\:to\: solve\: }}}$

<h3>➣Given</h3>

➣ (4h+5) and (h – 6) find the product

➣ Let's start !

$\qquad \sf{➛ \: (4h + 5) \times (h - 6)} \\$

★ Opening brackets multiplying nd solving

$\qquad \sf{➛( \: 4h \times h) +( 4h \times - 6) + (5 \times h) + (5 \times - 6)}\qquad \\ \sf{➛\: {4h}^{2} - 24h + 5h - 30 }$

★ arranging like terms and solving gives

$\qquad \sf{➛ \: {4h}^{2} - 19h - 30\bold\green{✓}}$

$\rule{70mm}{2.9pt}$

Hope it helps !

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The cosst of 4 1 /4 kg of sugar is £68 .find the coast o 1 kg

ankoles [38]

Step-by-step explanation:

Hi, there!!

Here according to the question we must find the cost of 1 kg sugar.

Given that:

17/4 kg of sugar cost £68.

then 1 kg sugar costs,

= $\frac{68}{ \frac{17}{4} }$ 

after reciprocal we get,

= £68×4/17

=£16

The answer would come £16.

Therefore, The cost of 1 kg sugar is £. 16.

Hope it helps....

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3 years ago

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Please help me with this question!!!!

Komok [63]

Answer:

EH = 13 in

Step-by-step explanation:

Angle between a tangent and radius is right, that is

∠ EBH = 90° and so Δ EBH is right

Using Pythagoras' identity in the right triangle

EH² = BE² + BH² = 5² + 12² = 25 + 144 = 169 ( square root both sides )

EH = $\sqrt{169}$ = 13

5 0

3 years ago

A thin metal plate, located in the xy-plane, has temperature T(x, y) at the point (x, y). Sketch some level curves (isothermals)

Sophie [7]

Answer:

Step-by-step explanation:

Given that:

$T(x,y) = \dfrac{100}{1+x^2+y^2}$

This implies that the level curves of a function(f) of two variables relates with the curves with equation f(x,y) = c

here c is the constant.

$c = \dfrac{100}{1+x^2+2y^2} \ \ \--- (1)$

By cross multiply

$c({1+x^2+2y^2}) = 100$

$1+x^2+2y^2 = \dfrac{100}{c}$

$x^2+2y^2 = \dfrac{100}{c} - 1 \ \ -- (2)$

From (2); let assume that the values of c > 0 likewise c < 100, then the interval can be expressed as 0 < c <100.

Now,

$\dfrac{(x)^2}{\dfrac{100}{c}-1 } + \dfrac{(y)^2}{\dfrac{50}{c}-\dfrac{1}{2} }=1$

This is the equation for the family of the eclipses centred at (0,0) is :

$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$

$a^2 = \dfrac{100}{c} -1 \ \ and \ \ b^2 = \dfrac{50}{c}- \dfrac{1}{2}$

Therefore; the level of the curves are all the eclipses with the major axis:

$a = \sqrt{\dfrac{100 }{c}-1}$ and a minor axis $b = \sqrt{\dfrac{50 }{c}-\dfrac{1}{2}}$ which satisfies the values for which 0< c < 100.

The sketch of the level curves can be see in the attached image below.

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3 years ago

2/3 of a plot is garden.1/5 of the garden is lawn find the fraction of the plot of land that is lawn

Marina CMI [18]

Answer:

2/3 + 1/5 = 13/15

Step-by-step explanation:

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3 years ago

suppose you write an equation that gives "a" as a function of "b". Which is the dependent variable and which is the independent

blondinia [14]

If im not wrong which i may not the
Dependent is A
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3 years ago

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