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Sedaia [141]
3 years ago
6

How do I solve x-11/2<-7

Mathematics
1 answer:
Mekhanik [1.2K]3 years ago
4 0

Answer:

x < -3

Step-by-step explanation:

1. multiply both sides by 2

2. isolate the x by adding 11 to both sides

3. simplify

(i have included an attachment showing the work for clarification)

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Keith_Richards [23]

Agree..............

3 0
3 years ago
N his history class, Han's homework scores are: 100% 100% 100% 100% 95% 100% 90% 100% 0% What is the mode of Han's scores?
Basile [38]

Answer:

There is no mode

Step-by-step explanation:

3 0
3 years ago
Given: KLMN is a trapezoid m∠N = m∠KML ME ⊥ KN , ME = 3 5 , KE = 8, LM KN = 3 5 Find: KM, LM, KN, Area of KLMN
prisoha [69]
A) Find KM∠KEM is a right angle hence ΔKEM is a right angled triangle Using Pythogoras' theorem where the square of hypotenuse is equal to the sum of the squares of the adjacent sides we can answer the 
KM² = KE² + ME²KM² = 8² + (3√5)²       = 64 + 9x5KM = √109KM = 10.44
b)Find LMThe ratio of LM:KN is 3:5 hence if we take the length of one unit as xlength of LM is 3xand the length of KN is 5x ∠K and ∠N are equal making it a isosceles trapezoid. A line from L that cuts KN perpendicularly at D makes KE = DN
KN = LM + 2x 2x = KE + DN2x = 8+8x = 8LM = 3x = 3*8 = 24
c)Find KN Since ∠K and ∠N are equal, when we take the 2 triangles KEM and LDN, they both have the same height ME = LD.
∠K = ∠N  Hence KE = DN the distance ED = LMhence KN = KE + ED + DN since ED = LM = 24and KE + DN = 16KN = 16 + 24 = 40
d)Find area KLMNArea of trapezium can be calculated using the  formula below Area = 1/2 x perpendicular height between parallel lines x (sum of the parallel sides)substituting values into the general equationArea = 1/2 * ME * (KN+ LM)          = 1/2 * 3√5 * (40 + 24)         = 1/2 * 3√5 * 64         = 3 x 2.23 * 32         = 214.66 units²
7 0
3 years ago
Assessment items What is the quotient? 1 5/8 ÷(−1 3/5 )
Solnce55 [7]

We need convert the mixed numbers to improper fractions.

1\dfrac{5}{8}=\dfrac{1\cdot8+5}{8}=\dfrac{8+5}{8}=\dfrac{13}{8}\\\\1\dfrac{3}{5}=\dfrac{1\cdot5+3}{5}=\dfrac{5+3}{5}=\dfrac{8}{5}

Divide the number by a fraction, this is the same as multiply this number by the reciprocal of the fraction:

1\dfrac{5}{8}\div\left(-1\dfrac{3}{5}\right)=-\dfrac{13}{8}\div\dfrac{8}{5}=-\dfrac{13}{8}\cdot\dfrac{5}{8}=-\dfrac{(13)(5)}{(8)(8)}=-\dfrac{65}{64}\\\\\boxed{1\dfrac{5}{8}\div\left(-1\dfrac{3}{5}\right)=-1\dfrac{1}{64}}

4 0
3 years ago
Rearrange y=1/2x+2 to make x the subject.
yarga [219]

Subtract 2

y-2=1/2 x

Multiply by 2 to cancel out the denominator and to get 1 in front of x

2(y-2)=x

You can stop there or you can multiply out.

2y-4=x


7 0
3 years ago
Read 2 more answers
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