Two shops, Tesco and Asda , sell the same brand of baked beans with the following deals.

Mathematics

1 answer:

Nikolay [14]2 years ago

5 0

Answer:

Asda offers a better value

Step-by-step explanation:

1.14 / 2 = 0.57

2.80 / 5 = 0.56

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Line segments XY and ZY are tangent to circle O. Circle O is shown. Triangle X Y Z has points X and Z on the circle. Lines X Y a

pickupchik [31]

Answer:

The correct option is;

An isosceles triangle

Step-by-step explanation:

The parameters given are;

Line segments XY and ZY are tangents to circle O.

ΔXYZ has points X and Z on the circle.

XY and ZY are tangents that intersect at point Y outside the circle'

Therefore, given the circle has center O, then;

OY ≅ OZ = Radius of circle with center O

∠OXY ≅ ∠OZY = 90° (The radius line from the circle center to the tangent point is perpendicular)

OY ≅ OY (Reflexive property)

ΔOYX and ΔOYZ are right triangles (Triangle with one angle = 90°

ΔOYX ≅ ΔOYZ (Hypotenuse Leg, HL, rule of congruency)

Therefore;

XY = ZY (Corresponding Parts of Congruent Triangles are Congruent CPCTC)

Triangle XYZ is an isosceles triangle (Triangle with two equal sides).

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2 years ago

Kevin wanted to go snowboarding for his vacation. Explain how he could make his decision regarding whether to go to Resort A or

Tcecarenko [31]

Answer:

Resort A has more consistent snowfall, so it shows less variation. However, the snowfall for Resort B has a higher median, and the interquartile range is higher (not larger), so it is more likely that Kevin will find a good snowfall at Resort B.

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Step-by-step explanation:

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2 years ago

Read 2 more answers

Find all the zeros of the equation 12x^2-64=-x^4

Natalija [7]

Answer: all the zeroes of the above equation will be

(x-4)(x+4)(x-2)(x+2)

Step-by-step explanation:

Since we have given that

$12x^2-64=-x^4\\\\x^4+12x^2-64=0$

We need to find the zeroes of the above equation.

So, we will use "Split the middle terms" :

$x^4+12x^2-64=0\\\\x^4+16x^2-4x^2-64=0\\\\x^2(x^2-16)-4(x^2-16)=0\\\\(x^2-4)(x^2-16)=0\\\\x^2-4=0\ or\ x^2-16=0\\\\\text{ "Using }a^2-b^2=(a+b)(a-b)"\\\\x^2-4=x^2-2^2=(x-2)(x+2)\\\\x^2-16=x^2-4^2=(x+4)(x-4)$