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3 years ago

The irrational number square root 67 is greater than [blank]

Mathematics

1 answer:

Tanya [424]3 years ago

3 0

Answer:

The irrational number square root 67 is greater than whole number 8

Step-by-step explanation:

Given

$\sqrt{67$

Required

Complete the sentence

No options are given.

So, we will solve the expression then compare it with the nearest smallest integer

First, we solve the given expression

$\sqrt{67$

$\sqrt{67} \approx 8.185$

By comparison:

$8.185 > 8$

This implies that, the blank will be completed with 8

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Let f (x)=x^2+2x-35 enter the x-intercepts of the quadratic function in the boxes __and___

serious [3.7K]

Answer:

The x-intercepts of the quadratic function are -7 and 5.

Step-by-step explanation:

The given function is

$f(x)=x^2+2x-35$

Equate f(x)=0, to find the x-intercepts of the function.

$x^2+2x-35=0$

$x^2+7x-5x-35=0$

$x(x+7)-5(x+7)=0$

$(x+7)(x-5)=0$

$x+7=0\Rightarrow x=-7$

$x-5=0\Rightarrow x=5$

Therefore the x-intercepts of the quadratic function are -7 and 5.

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gregori [183]

Answer:

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Step-by-step explanation:

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2 years ago

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horsena [70]

Area ( If you know the circumference) :
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3 years ago

For the order of operations, Group of answer choices Multiplication, division and remainder is performed first Addition and subt

Furkat [3]

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Step-by-step explanation:

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Answer:

Step-by-step explanation:

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