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4 years ago

Use co.patible numbers to find two estimates 73÷4,858

Mathematics

1 answer:

Svet_ta [14]4 years ago

6 0

NO.2 answer is 5088.052869

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What’s the constant 15-8y?

IgorC [24]

Answer:

Step-by-step explanation:

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3 years ago

Can sum1 do this for me

Ierofanga [76]

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3 years ago

H(x) = x2 1 k(x) = x – 2 (h k)(2) = (h – k)(3) = Evaluate 3h(2) 2k(3) =.

natima [27]

Quadratic equation is the equation in which only one variable is unknown. The highest power of the variable is 2.The value of the given functions are,

$(h+k)(x)=5$

$(h-k)(x)=9$

$3h(2)+2k(3)=17$

<h3>Given information-</h3>

The given function is,

$h(x)=x^2+1$

$k(x)=x-2$

<h3>Quadratic equation</h3>

Quadratic equation is the equation in which only one variable is unknown. The highest power of the variable is 2.

1) The value of the function (h+k)(2),

$(h+k)(x)=h(x)+k(x)$

$(h+k)(x)=x^2+1+x-2$

$(h+k)(2)=2^2+1+2-2$

$(h+k)(x)=5$

2)The value of the function (h-k)(3),

$(h-k)(x)=h(x)-k(x)$

$(h-k)(x)=x^2+1-x+2$

$(h-k)(3)=3^2+1-3+2$

$(h-k)(x)=9$

3) The value of the function 3h(2)+2k(3)

$3h(x)+2k(x)=3x^2+3+2x-2\times 2$

$3h(2)+2k(3)=3\times2^2+3+2\times2-2\times 2$

$3h(2)+2k(3)=17$

Hence the value of the given functions are,

$(h+k)(x)=5$

$(h-k)(x)=9$

$3h(2)+2k(3)=17$

Learn more about the quadratic equation here;

brainly.com/question/2263981

4 0

3 years ago

Find the t-value such that the area in the right tail is 0.2 with 5 degrees of freedom.

evablogger [386]

Answer:

0.920

Step-by-step explanation:

To calculate this, we proceed to the t-table

Using degree of freedom 5 and significance level of 0.2, the t-value is 0.920

3 0

3 years ago

7^3 times 7^-7 times 7

dusya [7]

Hello there! :)

<u><em>Answer: $\frac{1}{343}$ </em></u>

<u><em>*The answer must have a fraction.*</em></u>

Step-by-step explanation:

Lesson: Zero and Negative Exponents.

Distributive property: a(b+c)=ab+ac

First, you apply exponent rule.

$a^b*a^c=a^b^+^c$

$7^-^3$

$\frac{1}{7^3}=7^3=7*7*7=343$

Final answer: $\frac{1}{343}$

Hope this helps!

Thanks!

-Charlie

Have a great day!

4 0

3 years ago