All categories

3 years ago

'Twice a number n decreased by five'.

Mathematics

1 answer:

Lemur [1.5K]3 years ago

6 0

Answer:

2n-5

Step by step Explanation:

You must put 2 in front of "n" to represent the multiplication of the 2 numbers. Then put a subtraction sign and five the show the number is decreased by 5.

Hope this helps!

You might be interested in

-5y + 12 + 8y − 6 − 2 = ______ help plz

Sladkaya [172]

Answer:

(combine like terms)

-5y + 12 + 8y − 6 − 2

3y+12-8

3y+4

3 0

3 years ago

Savannah had $13. She earned more money babysitting She now has at least $20.

NARA [144]

13+x greater than or equal to 20

5 0

3 years ago

4+2(8²×4²)+6+4+6+6+6+2+2

mash [69]

Answer: =2084

4+2(82)(42)+6+4+6+6+6+2+2

=4+264(42)+6+4+6+6+6+2+2

=4+2(64)(16)+6+4+6+6+6+2+2

=4+(2)(1024)+6+4+6+6+6+2+2

=4+2048+6+4+6+6+6+2+2

=2052+6+4+6+6+6+2+2

=2058+4+6+6+6+2+2

=2062+6+6+6+2+2

=2068+6+6+2+2

=2074+6+2+2

=2080+2+2

=2082+2

=2084

4 0

3 years ago

Read 2 more answers

Combine and simplify the following radical expression <br><br> ASAP ASAP ASAP ASAP

Montano1993 [528]

Answer:

$\sqrt[3]{3}$

Step-by-step explanation:

Our expression is: $\frac{1}{3} \sqrt[3]{81}$ .

Let's focus on the cube root of 81 first. What's the prime factorisation of 81? It's simply: 3 * 3 * 3 * 3, or $3^3*3$ . Put this in for 81:

$\sqrt[3]{81} =\sqrt[3]{3^3*3}=\sqrt[3]{3^3} *\sqrt[3]{3}$

We know that the cube root of 3 cubed will cancel out to become 3, but the cube root of 3 cannot be further simplified, so we keep that. Our outcome is then:

$\sqrt[3]{3^3} *\sqrt[3]{3}=3\sqrt[3]{3}$

Now, let's multiply this by 1/3, as shown in the original problem:

$\frac{1}{3}* 3\sqrt[3]{3}=\sqrt[3]{3}$

Thus, the answer is $\sqrt[3]{3}$ .

<em>~ an aesthetics lover</em>

6 0

3 years ago

Which equation gives the closest estimate of 692 – 343?

Oliga [24]

Answer:

350

Step-by-step explanation:

Nearest hundred: 300

Nearest whole number: 350

Hope this helped!

6 0

3 years ago