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3 years ago

Evaluate u + xy, for u = 2, x = 9, and y = 6.

Mathematics

2 answers:

yawa3891 [41]3 years ago

8 0

U+xy when u=2, x=9, and y=6

2+9(6)
2+54
56

puteri [66]3 years ago

6 0

Replace all variables with their equivalents.

The expression u + xy now becomes 2 + (9)(6).

Do the order of operations.

2 + (9)(6)
2 + 54
56

So, the answer is 56.

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Directed line segment overline FG has endpoints F(4, 9) and G(- 8, - 9) .On the coordinate plane below , plot a point that parti

Thepotemich [5.8K]

Answer:

The coordinates of the point that makes the division in the given ratio is (0,3)

Step-by-step explanation:

Here, we want to find the point on the line segment that divides the line segment in the ratio 1:2

We simply use the internal division formula

That would be;

(x,y) = (nx1 + mx2)/(m + n) , (ny1 + my2)/(m + n)

m = 1 and n = 2

(x1,y1) = (4,9)

(x2,y2) = (-8,-9)

Substituting these values into the formula, we have;

2(4) + 1(-8)/(1 + 2) , 2(9) + 1(-9)/(2 + 1)

= (8-8)/3 , (18-9)/3

= (0,3)

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3 years ago

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ohaa [14]

Answer:

x =±sqrt(225) = ±15

Step-by-step explanation:

x^2 = 225

Take the square root of each side

sqrt(x^2) = ±sqrt(225)

x =±sqrt(225) = ±15

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4 years ago

Which terms in the expression are like terms? 7x + (-7) + 2/3 + (-7m)

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Anna35 [415]

Answer:

101/2

Step-by-step explanation:

7 0

4 years ago

Read 2 more answers

Helppp me plsssssssss<br><br>

Oliga [24]

Answer:

The class 35 - 40 has maximum frequency. So, it is the modal class.

From the given data,

$\sf \:\:\:\:\:\:\:\:\:\:x_{k}=35$
$\sf \:\:\:\:\:\:\:\:\:\:f_{k}=50$
$\sf \:\:\:\:\:\:\:\:\:\:f_{k-1}=34$
$\sf \:\:\:\:\:\:\:\:\:\:f_{k+1}=42$
$\sf \:\:\:\:\:\:\:\:\:\:h=5$

${\bf \:\: {By\:using\:the\: formula}} \\ \\$

$\:\dag\:{\small{\underline{\boxed{\sf {Mode,\:M_{o} =\sf\red{x_k + {\bigg(h \times \: \dfrac{ ( f_k - f_{k-1})}{ (2f_k - f_{k - 1} - f_{k +1})}\bigg)}}}}}}} \\ \\$

$\sf \:\:\:\:\:\:\:\:\:= 35+ {\bigg(5 \times \dfrac{(50 - 34)}{ ( 2 \times 50 - 34 - 42)}\bigg)} \\ \\$

$\sf \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:= 35 +{\bigg(5 \times \dfrac{16}{24}\bigg)} \\ \\$

$\sf \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:= {\bigg(35+\dfrac{10}{3}\bigg)} \\ \\$

$\sf \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:(35 + 3.33) =.38.33 \\ \\$

$\:\:\sf {Hence,}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\ \large{\underline{\mathcal{\gray{ mode\:=\:38.33}}}} \\ \\$

${\large{\frak{\pmb{\underline{Additional\: information }}}}}$

MODE

Most precisely, mode is that value of the variable at which the concentration of the data is maximum.

MODAL CLASS

In a frequency distribution the class having maximum frequency is called the modal class.

${\bf{\underline{Formula\:for\: calculating\:mode:}}} \\$

${\underline{\boxed{\sf {Mode,\:M_{o} =\sf\red{x_k + {\bigg(h \times \: \dfrac{ ( f_k - f_{k-1})}{ (2f_k - f_{k - 1} - f_{k +1})}\bigg)}}}}}} \\ \\$

Where,

$\sf \small\pink{ \bigstar} \: x_{k}= lower\:limit\:of\:the\:modal\:class\:interval.$

$\small \blue{ \bigstar}$ $\sf \: f_{k}=frequency\:of\:the\:modal\:class$

$\sf \small\orange{ \bigstar}\: f_{k-1}=frequency\:of\:the\:class\: preceding\:the\;modal\:class$

$\sf \small\green{ \bigstar}\: f_{k+1}=frequency\:of\:the\:class\: succeeding\:the\;modal\:class$

$\small \purple{ \bigstar}$ $\sf \: h= width \:of\:the\:class\:interval$

7 0

3 years ago